Similar Triangles

SIMILAR TRIANGLES (i) Any two square are similar (see fig. (i)) CONTENTS   Concept of Similarity   Thales Theorem   Criteria for Similarity of Triangles   Area of Two Similar Triangles   Phythagoras Theorem   Some Important Theorems Fig.(i) Fig.(ii) (ii) Any two equilateral triangles are similar (see fig. (ii)) SIMILAR POLYGONS  Definition CONCEPT OF SIMILARITY Geometric figures having the same shape but different sizes are known as similar figures. Two congruent figures are always similar but similar figures need not be congruent. Illustration 1 : Any two line segments are always similar but they need not be congruent. They are congruent, if their lengths are equal. Illustration 2 : Any two circles are similar but not necessarily congruent. They are congruent if their are equal. Two polygons are said to be similar to each other, if (i) their corresponding angles are equal, and (ii) the lengths of their corresponding sides are proportional. If two polygons ABCDE and PQRST are similar, then from the above definition it follows that : Angle at A = Angle at P, Angle at B = Angle at Q, Angle at C = Angle at R, Angle at D = Angle at S, Angle at E = Angle at T and, AB BC CD DE = = = PQ QR RS ST If two polygons ABCDE and PQRST, are similar, we write ABCED ~ PQRST. Here, the symbol ‘~’ stands for is similar to. Illustration 3 : A S cm 3c 120° 4 cm C P B R 90° 120° 6m Now, Area(ADE) = Two triangles are said to be similar, if their and, Area(DBE) = (i) corresponding angles are equal and, (ii) corresponding sides are proportional. Two triangles ABC and DEF are similar, if  (i) A = D, B = E, C = F and, AB BC AC = = DE EF DF  1 1 (base × height) = (AD.EF) 2 2 1 1 (base × height) = (DB.EF) 2 2 1 (AD.EF) Area (ADE) 2 AD = = .... (i) 1 Area (DBE) DB ( DB.BF) 2 Similarly, we have 1 (AE.DG ) Area (ADE) 2 AE = = 1 Area( DEC) (EC.DG ) EC 2 D A .... (ii) But, DBE and DEC are on the same base DE and between the same parallels DE and BC. C E F  Area (DBE) = Area (DEC) SOME BASIC RESULTS ON PROPORTIONALITY  Basic Proportionality Theorem or Thales Theorem If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.  AD AE = DB EC Construction : Join BE, CD and draw EF  BA and DG  CA.  1 1 = Area(DBE) Area(DEC)  [Taking reciprocals of both sides]   Given : A triangle ABC in which DE || BC, and intersects AB in D and AC in E. To Prove : C Proof : Since EF is perpendicular to AB. Therefore, EF is the height of triangles ADE and DBE. Definition B E B Q SIMILAR TRIANGLE AND THEIR PROPERTIES (ii) G F D m m 90° 100° m 3.2 A T m 4.8 c 3 130° 100° E 4c 6c 4.5 c D m cm m 5 c 130° 4. [Multiplying both sides by Area (ADE)]   Area (ADE) Area (ADE) = Area (DBE) Area (DEC) AD AE = DB EC [Using (i) and (ii)] Corollary : If in a ABC, a line DE || BC, intersects AB in D and AC in E, then : (i) AB AC = AD AE (ii) AB AC = EC DB Proof : (i) From the basic proportionality theorem, we have Proof : If possible, let DE be not parallel to BC. Then, there must be another line parallel to BC. Let DF || BC. AD AE = DB EC DB EC = [Taking reciprocals of both sides] AD AE   1+  Since DF || BC. Therefore Proportionality Theorem, we get DB EC =1+ [Adding 1 on both sides] AE AD AD AF = DB FC AD  DB AE  EC AB AC =  = AD AE AD AE  But, (ii) From the basic proportionality theorem, we have AD AE +1= +1 DB EC         So, if in a ABC, DE || BC, and intersect AB in D and AC in E, then we have (i) (iii) (v)  [Adding 1 on both sides] AD  DB AE  EC AB AC =  = DB EC DB EC AD AE = DB EC (ii) AB AC = AD AE (iv) AD AE = AB AC (vi) DB EC = AB AC AB AC = DB EC Basic .... (i) AD AE = DB EC (Given) .... (ii) From (i) and (ii), we get AF AE = FC EC AD DE = DB EC  from  AF AE +1= + 1 [Adding 1 on both sides] FC EC   AF  FC AE  EC = FC EC   AC AC = FC = EC FC EC This is possible only when F and E coincide i.e. DF is the line l itself. But, DF || BC. Hence, l || BC. DB EC = AD AE EXAMPLES  Ex.1 D and E are points on the sides AB and AC respectively of a ABC such that DE || BC. Find the value of x, when A Converse of Basic Proportionality Theorem If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. D B Given : A ABC and a line l intersecting AB in AD AE = D and AC in E, such that DB EC D B F E  C C (i) AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm (ii) AD = (7x – 4) cm, AE = (5x – 2) cm, To prove :  || BC i.e. DE || BC A E DB = (3x + 4) cm and EC = 3x cm. Sol. (i) In ABC, DE || BC    AD AE = (By thales theorem) DB EC 4 8 = x4 3x  19 4(3x – 19) = 8(x – 4) Ex.3 12x – 76 = 8x – 32 In fig., EF || AB || DC. Prove that A 4x = 44 AE BF = . ED FC B x = 17 (ii) E In ABC, DE || BC  AD AE = DB EC 7x  4 3x  4 = (By thales theorem) 5x  2 3x Sol. We have, EF || AB || DC 6x2 – 26x + 8 = 0  EP || DC 3x2 – 13x + 4 = 0 Thus, in ADC, we have (x – 4) (3x – 1) = 0 x = 4, 1/3 Sol. C D 21x2 – 12x = 15x2 – 6x + 20x – 8 Ex.2 F P EP || DC Therefore, by basic proportionality theorem, we have Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA meeting CA, BA in M, N respectively; MN meets BC produced in T, prove that TX2 = TB × TC. AE AP = ED PC .... (i) Again, EF || AB || DC In TXM, we have  FP || AB A Thus, in CAB, we have FP || BA M N T B X Therefore, by basic proportionality theorem, we have BF AP = FC PC C From (i) and (ii), we have XM || BN  TB TM = TX TN AE BF = ED FC .... (i) Ex.4 In TMC, we have XN || CM  TN TX = TC TM ....(ii) In figure, A = B and DE || BC. Prove that AD = BE C .... (ii) Sol. D From equations (i) and (ii), we get E TB TX = TX TC  TX2 = TB × TC A B A = B (given)  BC = AC ....(i) R D Q S (Sides opposite to equal angles are equal) Now, DE || AB  A CD CE = DA EB CD CE +1= +1 DA EB  PQ || AC ....(i) In DAC, S and R are the midpoints of AD and CD respectively.  CD  DA CE  EB CA CE =  = DA EB DA EB  SR || AC  AC BC AC AC =  = AD BE AD BE Similarly, PS || QR.  B Proof : In ABC, P and Q are the midpoints of AB and BC respectively. (Adding 1 on both sides) Ex.5 P Construction : Join AC. (By basic proportionality theorem)  C 1 1 = AD = BE AD BE In fig., DE || BC. If AD = 4x – 3, DB = 3x – 1, AE = 8x – 7 and EC = 5x – 3, find the value of x. ....(ii) From (i) and (ii), we get PQ || SR. Hence, PQRS is a parallelogram Ex.7 In fig. DE || BC and CD || EF. Prove that AD2 = AB × AF. Sol. In ABC, we have A C F E E D B A Sol. D DE || BC B In ABC, we have  DE || BC  AD AE = [By Thale’s Theorem] DB EC  4x  3 8x  7 = 3x  1 5x  3 Prove that the line segment joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram. Sol. Given : A quadrilateral ABCD in which P, Q, R, S are the midpoints of AB, BC, CD and DA respectively. To prove : PQRS is a parallelogram. ....(i) FE || DC  AD AC = AF AE ....(ii) From (i) and (ii), we get AB AD =  AD2 = AB × AF AD AF x=1 Ex.6 AB AC = AD AE In ADC, we have  (4x – 3) (5x – 3) = (3x – 1) (8x – 7)   C Ex.8 In the given figure PA, QB and RC each is perpendicular to AC such that PA = x, RC = y, QB = z, AB = a and BC = b. Prove 1 1 1 that + = . y x z R Q x y z a A  B  (x – 3) (x + 5) = (x – 2) (x + 3)  x2 + 2x – 15 = x2 + x – 6 b C PA  AC and QB  AC  QB || PA. Sol.  x=9 Thus, in PAC, QB || PA. So, QBC ~ PAC  QB BC z b =  = ....(i) PA AC x ab Ex.10 In a given ABC, DE || BC and AD 3 = . If DB 4 AC = 14 cm, find AE. [By the property of similar ] Sol. In ABC, we have DE || BC In RAC, QB || RC. So, QBC ~ RAC  x 3 x2 = x3 x5 z QB AB a =  = ....(ii) y RC AC ab  AD AE = DB EC [By Thales Theorem] [By the property of similar ] A From (i) and (ii), we get D z a  z  b + =   =1 y a  b a  b x   Ex.9 z z 1 1 1 + = 1  + = . x y x y z E B C 1 1 1 + Hence, = . y x z  AD AE = DB AC  AE In fig., LM || AB. If AL = x – 3, AC = 2x, BM = x – 2 and BC = 2x + 3, find the value of x.  3 AE = 4 14  AE  3(14 – AE) = 4AE C L [ AC = 5.6]  42 – 3AE = 4AE  42 = 7AE M  AE = 42 = 6 cm 7 Ex.11 In figure, DE || BC. Find AE. A C B B In ABC, we have Sol. LM || AB   9 cm E  AL MB = LC MC  AL BM = AC  AL BC  BM  x 3 x2 = 2x  ( x  3) (2x  3)  ( x  2) D 6 cm [By Thale’s Theorem]  A Sol. Let AE = x cm Then EC = (9 – x) cm AD = 2 cm DB = (6 – 2) cm = 4 cm We have Again, DE || BC AE AD =  BE DB  ACB = AED [By Basic Proportionality Theorem]  x 2 = 9x 4  6x = 18  ACB + CED = AED + CED [Adding CED on both sides] 4x = 2 (9 – x)  ACB + CED = 180º x=3 ABC + CED = 180º [ ABC = ACB] Hence, AE = 3 cm Ex.12 In figure, ABC is a triangle in which AB = AC. Points D and E are points on the sides AB and AC respectively such that AD = AE. Show that the points B, C, E and D are concyclic. Sol. In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that  ABC + CED = 180º and ACB + BDE = 180º. Thus, BDEC is quadrilateral such that ACB + BDE = 180º and    ABC + CED = 180º AD 1 AE 1 = and = . Using DB 3 AC 4 converse of basic proportionality theorem, prove that DBC Ex.13 In fig., A In ABC, we have D AB = AC and AD = AE E A D E B B C Sol.  AB – AD = AC – AE AE 1 = AC 4  AC AC = 4  –1=3 AE AE  AC  AE EC AE 1 =3 = 3  = AE AE EC 3  DB = EC Thus, we have AD = AE and DB = EC  AD AE = DB EC  DE || BC [By the converse of Thale’s Theorem] C Ex.14 Using basic proportionality theorem, prove that the lines drawn through the points of trisection of one side of a triangle parallel to another side trisect the third side. A Sol. P1  ABC = ADE [Corresponding angles] P2  ABC + BDE = ADE + BDE [Adding BDE on both sides]  ABC + BDE = 180º  ACB + BDE = 180º [ AB = AC  ABC = ACB] B Q1 l1 Q2 l2 C l1 || BC, l2 || BC and AP1 = P1P2 = P2B (given) = Therefore, by basic proportionality theorem, we have 1 AB. 3 PE PD = EQ DA To prove, AQ1 = Q1Q2 = Q2B = 1 AC. 3 In PAR, we have DF || AD 1 AB AQ1 AP1 = = 3 Proof AC AB AB  PD PF = DA FR AQ1 1 1 =  AQ1 = AC AC 3 3 PE PF = EQ FR  EF || QR A E B C We have, [By the converse of Thale’s theorem] [By the converse of Basic Proportionality Theorem] Ex.17 Two triangles ABC and DBC lie on the same side of the base BC. From a point P on BC, PQ || AB and PR || BD are drawn. They meet AC in Q and DC in R respectively. Prove that QR || AD. Sol. AD AE =  DE || BC DB EC ....(ii) From (i) and (ii), we have AD AE = and ADE DB EC = ACB. Prove that ABC is an isosceles triangle. D [Given] Therefore, by basic proportionality theorem, we have Ex.15 In the given figure, Sol. ....(i) Given : Two triangles ABC and DBC lie on the same side of the base BC. Points P, Q and R are points on BC, AC and CD respectively such that PR || BD and PQ || AB. A ADE = ABC (corresponding s) D But, ADE = ACB (given)  ABC = ACB. Q R So, AB = AC [sides opposite to equal angles] Hence, ABC is an isosceles triangles. Ex.16 In fig., if DE || AQ and DF || AR. Prove that EF || QR. [NCERT] Sol. In PQA, we have B Proof : In ABC, we have PQ || AB P  D CQ CP = QA PB ....(i) [By Basic Proportionality Theorem] F In BCD, we have A PR || BD Q C To Prove : QR || AD DE || AQ [Given] E P R  CP CR = PB RD From (i) and (ii), we get ....(ii) AE BF = ED FC [By Thale’s Theorem] From (i) and (ii), we have Ex.19 In fig., A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. [NCERT] CQ CR = QA RD Thus, in ACD, Q and R are points on AC and CD respectively such that Sol. In OPQ, we have AB || PQ CQ CR = QA RD P  QR || AD [By the converse of Basic proportionality theorem] A Ex.18 ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF || AB. Show that B D  C B  OA OC = AP CR  EF || DC Now, in ADC, EG || DC Similarly, in CAB, GF || AB. AG BF = GC FC [ ....(ii) GC FC = by Thale’s theorem] AG BF ....(ii) Thus, A and C are points on sides OP and OR respectively of OPR, such that Proof : EF || AB and AB || DC ....(i) [By Thale’s theorem] OB OC = BQ CR OA OC = AP CR Construction : Ioin AC, intersecting EF at G.  ....(i) From (i) and (ii), we get AE BF = ED FC AE AG  = ED GC OB OA = BQ AP BC || QR Given : A trap. ABCD in which AB || DC. E and F are points on AD and BC respectively such that EF || AB. To prove : R In OQR, we have F G A Sol. C Q AE BF = . ED FC E O AC || PR [Using the converse of BPT] Ex.20 Any point X inside DEF is joined to its vertices. From a point P in DX, PQ is drawn parallel to DE meeting XE at Q and QR is drawn parallel to EF meeting XF in R. Prove that PR || DF. Sol. A DEF and a point X inside it. Point X is joined to the vertices D, E and F. P is any point on DX. PQ || DE and QR || EF. To Prove : PR || DF E Construction : Join PR. 4 Proof : In XED, we have PQ || DE A D 12 P 3 B X Q E  F XQ XP = ....(i) [By Thale’s Theorem] QE PD QR || EF  2 = 3 From (i) and (ii), we have XP XR = PD RF Thus, in XFD, points R and P are dividing sides XF and XD in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have, PR || DF But, 1 = 2 Given : A ABC in which AD is the internal bisector of A and meets BC in D. BD AB = DC AC Construction : Draw CE || DA to meet BA produced in E. ....(ii) [ AD is the bisector of A] From (i) and (ii), we get 3 = 4 Thus, in ACE, we have 3 = 4  AE = AC ....(iii) [Sides opposite to equal angles are equal] Now, in BCE, we have DA || CE Theorem 1 : The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. [Alternate angles] ....(i) and,1 = 4 [Corresponding angles] XQ XR ...(ii) [By Thale’s Theorem] = QE RF To Prove : C Proof : Since CE || DA and AC cuts them. R In XEF, we have  D  BD BA = DC AE  BD AB = DC AC [ BA = AB and AE = AC (From (iii))] Hence, BD AB = DC AC Theorem 2 : The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle. k A E 4 B To Prove : E is the mid-point of AC i.e., AE = EC. 2 Proof : In ABC, we have 1 DE || BC 3  C D AD AE = [By Thale’s Theorem] ....(i) DB EC A Ex.21 Prove that any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally. Sol. Given : A trapezium ABCD in which DC || AB and EF is a line parallel to DC and AB. But, D is the mid-point of AB.  Construction : Join AC, meeting EF in G. E C G AD =1 DB AE =1  EC B EG || DC AE AG = [By Thale’s Theorem]....(i) ED GC In ABC, we have   AE = EC Hence, E bisects AC. Ex.23 Prove that the line joining the mid-point of two sides of a triangle is parallel to the third side. [NCERT] Sol. Given : A ABC in which D and E are midpoint of sides AB and AC respectively. A GF || AB  ....(ii) From (i) and (ii), we get F Proof : In ADC, we have   AD = DB   A  C B AE BF To Prove : = ED FC D E D AG BF = [By Thale’s Theorem]....(ii) GC FC D E From (i) and (ii), we get AE BF = ED FC Ex.22 Prove that the line drawn from the mid-point of one side of a triangle parallel of another side bisects the third side. Sol. Given : A ABC, in which D is the midpoint of side AB and the line DE is drawn parallel to BC, meeting AC in E. B C To Prove : DE || BC Proof : Since D and E are mid-points of AB and AC respectively.  AD = DB and AE = EC  AE AE = 1 and =1 DB EC  AD AE = DB EC Ex.25 O is any point inside a triangle ABC. The bisector of AOB, BOC and COA meet the sides AB, BC and CA in point D, E and F respectively. Show that AD × BE × CF = DB × EC × FA. Thus, the line DE divides the sides AB and AC of ABC in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have Sol. In AOB, OD is the bisector of AOB. A DE || BC F D Ex.24 AD is a median of ABC. The bisector of ADB and ADC meet AB and AC in E and F respectively. Prove that EF || BC. Sol. Given : In ABC, AD is the median and DE and DF are the bisectors of ADB and ADC respectively, meeting AB and AC in E and F respectively. To Prove : EF || BC Proof : In ADB, DE is the bisector of ADB. A O B  B  AD AE = DB EB ....(i)  OB BE = OC EC ....(ii) In COA, OF is the bisector of COA OC CF = OA FA ....(iii) OA OB OC AD BE CF × × = × × OB OC OA DB EC FA C ....(i) In ADC, DF is the bisector of ADC.  AD AF = DC FC  AD AF  AD is the median =   ....(ii) DB FC  BD  DC  From (i) and (ii), we get AE AF = EB FC Thus, in ABC, line segment EF divides the sides AB and AC in the same ratio. Hence, EF is parallel to BC. OA AD = OB DB Multiplying the corresponding sides of (i), (ii) and (iii), we get F D C In BOC, OE is the bisector of BOC.  E E  1= AD BE CF × × DB EC FA  DB × EC × FA = AD × BE × CF  AD × BE × CF = DB × EC × FA = 180º – 50º – 70º = 60º CRITERIA FOR SIMILARITY OF TRIANGLES  L = 60º   Equiangular Triangles : Two triangles are said to be equiangular, if their corresponding angles are equal. If two triangles are equiangular, then they are similar. Two triangles ABC and DEF such that    A = D, B = E and C = F. Then ABC ~ DEF and AB BC AC = = DE EF DF A D Ex.27 Examine each pair of triangles in Figure, and state which pair of triangles are similar. Also, state the similarity criterion used by you for answering the question and write the similarity relation in symbolic form. A P 50º 60º 70º 60º B B C E C F 70º 50º Q R Figure (i) D  EXAMPLES  Ex.26 In figure, find L. P M A B 4 cm N C 6 cm 3 cm 2.5 cm 70° 50° 5 cm Q 2 cm R E Figure (ii) E L A In ABC and LMN, 70° B CA 3.6 2 and = = NL 9 5  4.5 cm C Figure (iii) AB BC CA = = LM MN NL  ABC ~ LMN m BC 4 2 = = MN 10 5 70° 9c AB 4.4 2 = = LM 11 5 3c m Sol. F 4 cm (SSS similarity)  L = A = 180º – B – C F D P cm (iv) CAB ~ QRP (SAS-similarity), as 4.5 80° m 6c 8c m C (iii) SAS-similarity is not satisfied as included angles are not equal. 6 cm CA CB = and C = Q. QR QP 80° Q 6c m (v) In ’s ABC and DEF, we have B A R A = D = 80º Figure (iv) 80° B 3 cm F 6 cm D D = M = 70º Figure (v) D M 70° 70° E = N = 80º [ N = 180º – M – P = 180º – 70º – 30º = 80º] So, by AA-criterion of similarity 80° 30° F E N DEF ~ MNP. P (vii) FE = 2 cm, FD = 3 cm, ED = 2.5 cm Figure (vi) D [ AC is not given] (vi) In ’s DEF and MNP, we have 80° C AB AC  DE DF So, by SAS-criterion of similarity these two triangles are not similar. 5 cm 2.5 cm But, E A PQ = 4 cm, PR = 6 cm, QR = 5 cm P 2 cm 6 4c m 2.5 E m 3c cm  FED ~ PQR (SSS-similarity) F Q Ex.28 In figure, QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Find AQ. cm 5 cm R Figure (vii) Sol. (i) A = Q, B = P and C = R. ABC ~ QPR (AAA-similarity) (ii) In triangle PQR and DEF, we observe that PQ QR PR 1 = = = DE EF DF 2 Sol. In triangles AOQ and BOP, we have OAQ = OBP AOQ = BOP [Vertically opposite angles] Therefore, by AA-criterion of similarity AOQ ~ BOP P A Therefore, by SSS-criterion of similarity, we have PQR ~ DEF [Each equal to 90º] O B Q  AO OQ AO = = BO OP BP  AO AQ 10 AQ =  = BO BP 6 9  AQ = Ex.31 In figure, CAB = 90º and AD  BC. If AC = 75 cm, AB = 1 m and BD = 1.25 m, find AD. Sol. AB = 1 m = 100 cm, AC = 75 cm and BD = 125 cm 10  9 = 15 cm 6 In BAC and BDA, we have Ex.29 In figure, ACB ~ APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, find CA and AQ. Sol. We have, BAC = BDA B = B and, We have, ACB ~ APQ [Each equal to 90º] C D 75 cm P A Q D 1. 2 A C CB AB AC = = PQ AQ AP  CB CB AB AC and = = PQ PQ AQ AP   6.5 AC 8 8 = and = AQ 2.8 4 4   6.5 AC =2 = 2 and AQ 2.8  AD = So, by AA-criterion of similarity, we have BAC ~ BDA  AC = (2 × 2.8) cm = 5.6 cm and 6.5 cm = 3.25 cm AQ = 2 BA AC = BD AD 100 75 = 125 AD 125  75 cm = 93.75 cm 100 Ex.32 In figure, if A = C, then prove that AOB ~ COD. A Ex.30 The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB. Sol. O 2  ABC ~ PQR D  AB BC AC 36 = = = PQ QR PR 24  AB 36 = PQ 24  AB = AB 36 = 10 24 36  10 cm = 15 cm 24 B 1 Since the ratio of the corresponding sides of similar triangles is same as the ratio of their perimeters.  B 1m  5m Sol. C In triangles AOB and COD, we have A = C [Given] and,1 = 2 [Vertically opposite angles] Therefore, by AA-criterion of similarity, we have AOB ~ COD AO BO 1 = = and AB = 5 cm. OC OD 2 Find the value of DC. Ex.33 In figure, Sol. In AOB and COD, we have AOB = COD [Vertically opposite angles]  BP × PD = EP × PC Ex.35 D is a point on the side BC of ABC such CA CB that ADC = BAC. Prove that = CD CA or, CA2 = CB × CD. Sol. AO OB = OC OD In ABC and DAC, we have [Given] 5 cm A ADC = BAC and C = C A B O D B C Therefore, by AA-criterion of similarity, we have So, by SAS-criterion of similarity, we have AOB ~ COD  AO BO AB = = OC OD DC  1 5 = 2 DC ABC ~ DAC [ AB = 5 cm]  DC = 10 cm Ex.34 In figure, considering triangles BEP and CPD, prove that BP × PD = EP × PC. Sol. C D Given : A ABC in which BD  AC and CE  AB and BD and CE intersect at P. To Prove : BP × PD = EP × PC  AB BC AC = = DA AC DC  CB CA = CA CD Ex.36 P and Q are points on sides AB and AC respectively of ABC. If AP = 3 cm, PB = 6cm. AQ = 5 cm and QC = 10 cm, show that BC = 3PQ. Sol. We have, Proof : In EPB and DPC, we have AB = AP + PB = (3 + 6) cm = 9 cm PEB = PDC [Each equal to 90º] and, AC = AQ + QC = (5 + 10) cm = 15 cm. EPB = DPC [Vertically opposite angles]  AP 3 1 AQ 5 1 = = and = = AB 9 3 AC 15 3  AP AQ = AB AC A E D A P B P C Q Thus, by AA-criterion of similarity, we have EPB ~ DPC EP PB = DP PC B C Thus, in triangles APQ and ABC, we have AP AQ = and A = A [Common] AB AC Sol. Therefore, by SAS-criterion of similarity, we have To prove : ABD ~ ECF. APQ ~ ABC Proof : we known that the angles opposite to equal sides of a triangle are equal.  AP PQ AQ = = AB BC AC  PQ AQ PQ 5 =  = BC AC BC 15  PQ 1 =  BC = 3PQ BC 3  B = C [ AB = AC] Now, in ABD and ECF, we have B = C [proved above] ADB = EFC = 90º Ex.37 In figure, A = CED, prove that  CAB ~ CED. Also, find the value of x. 8 cm D Given : A ABC in which AB = AC and AD  BC. Side CB is produced to E and EF  AC. C  ABD ~ ECF [By AA-similarity] Ex.39 In figure, BAC = 90º and segment ADBC. Prove that AD2 = BD × DC. Sol. In ABD and ACD, we have A 10 cm x 7 cm E 2 cm A B 9 cm B In CAB and CED, we have Sol. ADB = ADC A = CED and C = C [common] Each equal to complement of  BAD i.e., 90º  BAD   CA AB CB = =  CE DE CD AB CB 9 10  2  =  = DE CD x 8      x = 6 cm  Ex.38 In the figure, E is a point on side CB produced of an isosceles ABC with AB = AC. If AD  BC and EF  AC, prove that    ABD ~ ECF. A B D Therefore, by AA-criterion of similarity, we have DBA ~ DAC  D  D, DBA  DAC   and BAD  DCA    DB DA =  DA DC In similar triangles corresponding  sides are proportional   F E [Each equal to 90º] and, DBA = DAC  CAB ~ CED  C D C BD AD = AD DC AD2 = BD × DC Ex.40 In an isosceles ABC, the base AB is produced both ways in P and Q such that AP × BQ = AC2 and CE are the altitudes. Prove that ACP ~ BCQ. FB FD = FD FA  C  FB EF = DF FA  DF × EF = FB × FA Ex.42 Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2 BL. P A Q B In BMC and EMD, we have CA = CB  CAB = CBA Sol.  MC = MD [ M is the mid-point of CD]  180º – CAB = 180º – CBA CMB = EMD [Vertically opposite angles]  CAP = CBQ and,MBC = MED Now, AP × BQ = AC2  Sol.  AC BC AP AP  =  = [ AC = BC] BQ BQ AC AC C [Alternate angles] So, by AAS-criterion of congruence, we have  EMD  BMC ~  BC = DE ....(i) Also, AD = BC ....(ii) [ ABCD is a parallelogram] AD + DE = BC + BC P A Q B Thus, CAP = CBQ and BC AP = . BQ AC ACP ~ BCQ. Ex.41 The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Prove that DF × EF = FB × FA. Sol.  AE = 2 BC Now, in AEL and CBL, we have ALE = CLB [Vertically opposite angles] EAL = BCL [Alternate angles] So, by AA-criterion of similarity of triangles, we have A In AFD and BFE, we have 1 = 2 [Vertically opposite angles] 3 = 4 [Alternate angles] 4 1 F2 A FBE ~ FDA AEL ~ CBL  EL AE EL 2BC =  = BL CB BL BC B So, by AA-criterion of similarity, we have M C B E 3 E D L C D ....(iii) [Using equations (iii)]  EL =2 BL  EL = 2BL Ex.43 In figure, ABCD is a trapezium with AB || DC. If AED is similar to BEC, prove that AD = BC. Sol. Sol. Let the sun’s altitude at that moment be . P In EDC and EBA, we have 1 = 2 [Alternate angles] 3 = 4 [Alternate angles] A h and, CED =AEB[Vertically opposite angles]  EDC ~ EBA A B B 2 4 D  ED EC = EB EA  ED EB = EC EA 3  MP AC = MQ CB  h 20 = 15 6 C Q 15 m M  Height of the tower = 50 m. ....(i) It is given that AED ~ BEC   PQM ~ ABC E 1  6 cm C ED EA AD = = EC EB BC ....(ii) From (i) and (ii), we get EB EA = EA EB Ex.45 If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to each other and to the original triangle. Also, prove that the square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse. [NCERT] Sol. Given : A right triangle ABC right angled at B, BD  AC.  (EB)2 = (EA)2 To Prove :  EB = EA (i) ADB ~ BDC (ii) ADB ~ ABC Substituting EB = EA in (ii), we get (iii) BDC ~ ABC (iv) BD2 = AD × DC EA AD AD =  =1 EA BC BC (v) AB2 = AD × AC (vi) BC2 = CD × AC AD = BC Proof : A Ex.44 A vertical stick 20 cm long casts a shadow 6 cm long on the ground. At the same time, a tower casts a shadow 15 m long on the ground. Find the height of the tower. D C B (i) We have, ABD + DBC = 90º Also, C + DBC + BDC = 180º  C + DBC + 90º = 180º But, ABD + DBC = 90º Ex.46 Prove that the line segments joining the mid points of the sides of a triangle form four triangles, each of which is similar to the original triangle.  ABD + DBC = C + DBC Sol.  C + DBC = 90º ....(i) Given :ABC in which D, E, F are the mid-points of sides BC, CA and AB respectively. [From (i)] To Prove : Each of the triangles AFE, FBD, EDC and DEF is similar to ABC.  ABD = C Thus, in ADB and BDC, we have ABD = C Proof : Consider triangles AFE and ABC. and,ADB = BDC [Each equal to 90º] So, by AA-similarity criterion, we have  FE || BC ADB ~ BDC  AEF = B (ii) In ADB and ABC, we have ADB = ABC Since F and E are mid-points of AB and AC respectively. [Corresponding angles] [Each equal to 90º] and,A = A [Common] Thus, in AFE and ABC, we have A So, by AA-similarity criterion, we have ADB ~ ABC BDC = ABC [Each equal to 90º] C = C B [Common] So, by AA-similarity criterion, we have BDC ~ ABC (iv) From (i), we have ADB ~ BDC  AD BD = BD2 = AD × DC BD DC (v) From (ii), we have ADB ~ ABC  AD AB = AB2 = AD × AC AB AC (vi) From (iii), we have BDC ~ ABC  BC DC = AC BC BC2 = CD × AC E F (iii) In BDC and ABC, we have D C AFE = B and,A = A [Common]  AFE ~ ABC. Similarly, we have FBD ~ ABC and EDC ~ ABC. Now, we shall show that DEF ~ ABC. Clearly, ED || AF and DE || EA.  AFDE is a parallelogram.  EDF = A [ Opposite angles of a parallelogram are equal] Similarly, BDEF is a parallelogram.  DEF = B [ Opposite angles of a parallelogram are equal] Thus, in triangles DEF and ABC, we have EDF = A and DEF = B BC 5 = DE 2 So, by AA-criterion of similarity, we have DEF ~ ABC.  MORE ON CHARACTERISTIC PROPERTIES Thus, each one of the triangles AFE, FBD, EDC and DEF is similar to ABC. Ex.47 In ABC, DE is parallel to base BC, with D AD 2 BC on AB and E on AC. If = , find . DB 3 DE In ABC, we have Sol. A Theorem 1 : If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding medians. Given : Two triangles ABC and DEF in which A = D, B = E, C = F, AP and DQ are their medians. D B DE || BC A E D C  AB AC = AD AE To Prove : Therefore, by SAS-criterion of similarity, we have ABC ~ ADE AD BC = AD DE   ....(i)  DB 3 = AD 2  DB 3 +1= +1 AD 2  DB  AD 5 = AD 2  AB 5 = DE 2 From (i) and (ii), we get F Figure (ii) AP BC = DQ EF  ABC ~ DEF   Q E Proof : Since equiangular triangles are similar. 2BP AB BC AB =  = 2EQ DE EF DE  P and Q are mid  po int s of BC  and EF respectively    BC  2BP and EF  2EQ  It is given that AD 2 = DB 3 C Figure (i) Thus, in triangles ABC and ADE, we have  P B AB AC = and, A = A AD AE   BP AB = EQ DE ....(ii) Now, in ABP and DFQ, we have BP AB = EQ DE and, B = E [From (ii)] [Given] So, by SAS-criterion of similarity, we have ABP ~ DEQ ....(ii)  AP AB = DQ DE From (i) and (iii), we get ....(iii) AP BC = DQ EF BC AX = EF DY Theorem 3 : Hence, the ratio of the corresponding sides is same as the ratio of corresponding medians. If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding altitudes. Theorem 2 : If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments. Given : Two triangles ABC and DEF in which  Given : Two triangles ABC and DEF in which A = D, B = E, C = F; and AX, DY are the bisectors of A and D respectively. A = D, B = E, C = F and AL  BC, DM  EF D A D A B B X C E Figure (i) Y F To Prove : Figure (ii)  Proof : Since equiangular triangles are similar. ....(i) [Given] 1 1   A  D  2 A  2 D   BAX  EDY  AB BC = DE EF ....(i) ALB = DME [Each equal to 90º] [Given] So, by AA-criterion of similarity, we have ALB ~ DME    So, by AA-criterion of similarity, we have  AB AL = DE DM ....(ii) From (i) and (ii), we get BC AL = EF DM ABX ~ DEY From (i) and (ii), we get BC AL = EF DM B = E and,BAX = EDY AB AX = DE DY F In triangle ALB and DME, we have AB BC = DE EF In ABX and DEY, we have  M  ABC ~ DEF ABC ~ DEF B = E E Proof : Since equiangular triangles are similar. BC AX To Prove : = EF DY  C L ....(ii) Theorem 4 : If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite side in the same ratio, prove that the triangles are similar. the two sides and the corresponding median of another triangle, then the triangles are similar. Given : Two triangles ABC and DEF in which A = D. The bisectors AP and DQ or A and D intersect BC and EF in P and Q respectively EQ BP such that = . QF PC Given : ABC and DEF in which AP and DQ are the medians such that [NCERT] AP AB BC = = DQ DE EF To Prove : ABC ~ DEF Proof : We know that the bisectors of an angle of triangle of a triangle intersects the opposite side in the ratio of the sides containing the angle. To Prove : ABC ~ DEF Proof : We have, AP AB BC = = DQ DE EF D A D A B P C E Q F  AP is the bisector of A  B BP AB = PC AC C E 1 BC AP AB = 2 =  1 DQ DE EF 2 ....(i) DQ is the bisector of D   But, EQ DE = QF DF EQ BP = QF PC ....(ii) [Given] Therefore, from (i) and (ii), we get F BP AP AB = = EQ DQ DE     ABP ~ DEQ   B = E AB BC = DE EF Thus, in triangles ABC and DEF, we have AB DE = AC DF A = D Q [By SSS-similarity] Now, in ABC and DEF, we have AB DE = AC DF and, P [Given] and, B = E So, by SAS-criterion of similarity, we get ABC ~ DEF [Given] So, by SAS-criterion of similarity, we get ABC ~ DEF Theorem 5 : If two sides and a median bisecting one of these sides of a triangle are respectively proportional to Theorem 6 : If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then the two triangles are similar. [NCERT] Given : Two triangle ABC and DEF in which AP and DQ are the medians such that AP AB AC = = . DQ DE DF  To Prove : ABC ~ DEF  AGC ~ DHF Construction : Produce AP to G so that PG = AP. Join CG. Also, produce DQ to H so that QH = DQ. Join FH. [By SSS-criterion of similarity]  1 = 2 Proof : In APB and GPC, we have BP = CP AP = GP Similarly, we have [ AP is the median] [By construction]   A = D So, by SAS-criterion of congruence, we have  ....(i) D and, 4 2 A = D E So, by SAS-criterion of similarity, we have F Q  AREAS OF TWO SIMILAR TRIANGLES Theorem 1 : C P G The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides. H Again, In DQE and HQF, we have EQ = FQ [ DQ is the median] DQ = HQ [By construction] Given : Two triangles ABC and DEF such that ABC ~ DEF. To Prove : and, DQE = HQF [Vertically opposite angles]  DE = HF AP AB AC = = DQ DE DF A ....(ii) [Given] B  2AP GC AC = = HF DF 2DQ L C E M F Construction : Draw AL  BC and DM  EF. AP GC AC = = DQ HF DF Proof : Since similar triangles are equiangular and their corresponding sides are proportional. Therefore, [ AB = GC and DE = HF (from (i) and (ii))]  Area (ABC) AB2 BC 2 AC 2 = = = Area (DEF) DE 2 EF 2 DF 2 D So, by SAS-criterion of congruence, we have   DQE ~  HQF  [Given] ABC ~ DEF 31 Now, [From (iii)] AB AC = DE DF A B ....(iii) Thus, in ABC and DEF, we have APB  GPC  AG = GC 3 = 4  1 + 3 = 2 + 4 and, APB = CPG [Vertically opposite angles]  GC AC AG = = HF DF DH [ 2AP = AG and 2DQ = DH]   ABC ~ DEF  A = D, B = E, C = F AB BC AC = = DE EF DF and ABC ~  DEF ....(i) Thus, in ALB and DME, we have  ALB = DME Proof : ABC ~ DEF  [Each equal to 90º] and, B = E [From (i)] So, by AA-criterion of similarity, we have ALB ~ DME  AL AB = DM DE  Now, ....(ii)  From (i) and (ii), we get AB BC AC AL = = = DE EF DF DM  A = D, B = E, C = F and AB BC AC = = DE EF DF In order to prove that ABC ~  DEF, it is sufficient to show that AB = DE, BC = EF and AC = DF. ....(iii) Area (ABC) = Area (DEF)  Area ( ABC) =1 Area (DEF)  AB2 BC 2 AC 2 = = =1 DE 2 EF 2 DF 2  Area (ABC) AB2 BC 2 AC 2       2 EF 2 DF 2   Area ( DEF) DE Now, 1 (BC  AL) Area ( ABC) = 2 1 Area (DEF) (EF  DM ) 2      Area ( ABC) BC AL = × Area (DEF) EF DM  Area(ABC) BC BC = × Area(DEF) EF EF BC AL   From(iii), EF  DM     AB2 = DE2, BC2 = EF2 and AC2 = DF2   AB = DE, BC = EF and AC = DF Hence, ABC  DEF. EXAMPLES  Ex.48 The areas of two similar triangles ABC and PQR are 25 cm2 and 49 cm2 respectively. If QR = 9.8 cm, find BC. It is being given that ABC ~ PQR, ar (ABC) = 25 cm2 and ar (PQR) = 49 cm2. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Sol. 2  Area ( ABC) BC = Area (DEF) EF 2 But, BC AB AC = = EF DE DF  BC 2 AB2 AC 2 = = EF 2 DE 2 DF 2 P A B Theorem 2 : If the areas of two similar triangles are equal, then the triangles are congruent i.e. equal and similar triangles are congruent. Given : Two triangles ABC and DEF such that  ABC ~ DEF and Area (ABC) = Area (DEF). To Prove : We have, 49 cm2 25 cm2 Area (ABC) AB2 BC 2 AC 2 Hence, = = = Area (DEF) DE 2 EF 2 DF 2 x cm C Q 9.8 cm   ar (ABC) BC 2 = ar (PQR ) QR 2   x2 25 , where BC = x cm = 49 (9.8) 2  25   x2 =   9.8  9.8   49  R  x= 25 5   9.8  9.8 =   9.8  = (5 × 1.4) = 7. 49 7   Hence BC = 7 cm. AB = 2 × CD. If the area of AOB = 84 cm2. Find the area of COD. Sol. In AOB and COD, we have D Ex.49 In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS are in the ratio 4 : 9, find the ratio of the areas of ABC and PQR. Sol. O Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.   A  B OAB = OCD (alt. int. s) 2 Area ( ABC) AD = Area (PQR ) PS 2 OBA = ODC (alt. int. s)  AOB ~ COD [By AA-similarity] 2  C Area (ABC) 16 4 =  = Area (PQR ) 81 9  ar (AOB) AB2 (2CD) 2 = = ar (COD) CD 2 CD 2 [ AD : PS = 4 : 9] [ AB = 2 × CD] Hence, Area (ABC) : Area (PQR) = 16 : 81 = Ex.50 If ABC is similar to DEF such that  DEF = 64 cm2, DE = 5.1 cm and area of ABC = 9 cm2. Determine the area of AB. Sol.   AB2 9 = 64 (5.1) 2  AB = 1  =   84  cm2 = 21cm2 4  3.65 AB = 1.912 cm We have, (2.3) 2 16  EF = = 25 EF 2 Sol. ABCD is a square. BCE is described on side BC is similar to ACF described on diagonal AC. Since ABCD is a square. Therefore, AB = BC = CD = DA and, AC = [ Diagonal = Area ( ABC) BC 2 = Area (DEF) EF 2  Hence, the area of COD is 21 cm2. Ex.53 Prove that the area of the triangle BCE described on one side BC of a square ABCD as base is one half the area of the similar triangle ACF described on the diagonal AC as base. Ex.51 If ABC ~ DEF such that area of ABC is 16cm2 and the area of DEF is 25cm2 and BC = 2.3 cm. Find the length of EF. Sol.  ar (COD) = 1/4 × ar (AOB) Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Area ( ABC) AB2 = Area (DEF) DE 2 4  CD 2 =4 CD 2 2 BC 2 (Side)] F 8.265 D C E = 2.875 cm Ex.52 In a trapezium ABCD, O is the point of intersection of AC and BD, AB || CD and A Now, BCE ~ ACF B  Area ( BCE) BC 2 = Area ( ACF) AC 2  Area ( BCE) BC 2 1 = = 2 Area ( ACF) 2 ( 2BC)  Area (BCE) = and divides ABC into two parts, equal in BD area. Find . AB Sol. We have, Area (ADE) = Area (trapezium BCED)  Area (ADE) + Area (ADE) 1 Area (ACF) 2 = Area (trapezium BCED) + Area (ADE) Ex.54 D, E, F are the mid-point of the sides BC, CA and AB respectively of a ABC. Determine the ratio of the areas of DEF and ABC.  2 Area (ADE) = Area (ABC) Sol. [ DE || BC  ADE = B (Corresponding angles)] In ADE and ABC, we have ADE = B Since D and E are the mid-points of the sides BC and AB respectively of ABC. Therefore, DE || BA  DE || FA and, A = A ....(i)  ADE ~ ABC A B A D E F [Common] D E C B C Since D and F are mid-points of the sides BC and AB respectively of ABC. Therefore,  Area ( ADE) AD 2 = Area (ABC) AB2  AD 2 Area (ADE) = 2 Area ( ADE) AB2  1 1  AD  AD = =   2  AB  AB 2 DF || CA  DF || AE From (i), and (ii), we conclude that AFDE is a parallelogram. Similarly, BDEF is a parallelogram. 2 Now, in DEF and ABC, we have FDE = A  AB = [Opposite angles of parallelogram AFDE] 2 AD AB =  ( 2 – 1) AB = and, DEF = B [Opposite angles of parallelogram BDEF]  So, by AA-similarity criterion, we have DEF ~ ABC Area (DEF) DE 2 (1 / 2AB) 2 1  = = = Area ( ABC) AB2 4 AB2 1    DE  2 AB   Hence, Area (DEF) : Area (ABC) = 1 : 4. Ex.55 D and E are points on the sides AB and AC respectively of a ABC such that DE || BC BD = AB 2 1 2 2 (AB – BD) 2 BD = 2 2 2 Ex.56 Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights. Sol. Let ABC and DEF be the given triangles such that AB = AC and DE = DF, A = D. and, Area ( ABC) 16 = Area (DEF) 25 Draw AL  BC and DM  EF. ....(i) Now, AB = AC, DE = DF  ar (ADE) 9x 9 = = ar ( trap.BCED) 16x 16  AB DE = 1 and =1 AC DF A  PYTHAGORAS THEOREM  D  Theorem 1 : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. L B  C AB DE = AC DF  M E Given : A right-angled triangle ABC in which B = 90º. F AB AC = DE DF To Prove : (Hypotenuse)2 = (Base)2 + (Perpendicular)2. i.e., AC2 = AB2 + BC2 Thus, in triangles ABC and DEF, we have AB AC = and A = D DE DF Construction : From B draw BD  AC. [Given] B So, by SAS-similarity criterion, we have ABC ~ DEF  AL2 Area ( ABC) = Area (DEF) DM 2 A C Proof : In triangle ADB and ABC, we have 16 AL2 =  25 DM 2  D [Using (i)] AL 4 =  AL : DM = 4 : 5 DM 5 ADB = ABC [Each equal to 90º] and,A = A [Common] So, by AA-similarity criterion, we have Ex.57 In the given figure, DE || BC and DE : BC = 3 : 5. Calculate the ratio of the areas of ADE and the trapezium BCED. ADB ~ ABC   AD AB = AB AC A  In similar triangles correspond ing   sides are proportion al   E D  AB2 = AD × AC ....(i) In triangles BDC and ABC, we have C B Sol. CDB = ABC ADE ~ ABC.  and, 2 2 2 ar (ADE) DE 9  DE   3 = =  =  = 2 ar (ABC) BC 25  BC  5 Let ar (ADE) = 9x sq units Then, ar (ABC) = 25x sq units ar (trap. BCED) = ar (ABC) – ar (ADE) = (25x – 9x) = (16x) sq units [Each equal to 90º] C = C [Common] So, by AA-similarity criterion, we have BDC ~ ABC   DC BC = BC AC  In similar triangles corresponding  sides are proportional   BC2 = AC × DC ....(ii)  ABC ~  DEF  AB2 + BC2 = AD × AC + AC × DC  B = E = 90º AB2 Hence, ABC is a right triangle right angled at B. + BC2 = AC (AD + DC)  AB2 + BC2 = AC × AC  AB2 + BC2 = EXAMPLES  AC2 Ex.58 Side of a triangle is given, determine it is a right triangle. Hence, AC2 = AB2 + BC2 The converse of the above theorem is also true as proved below. Theorem 2 : (Converse of Pythagoras Theorem). (2a – 1) cm, 2 2a cm and (2a + 1) cm Sol. r = (2a + 1) cm. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the side is a right angle. Then, (p2 + q2) = (2a – 1)2 cm2 + (2 2a )2 cm2 Given : A triangle ABC such that AC2 = AB2 + BC2 A Let p = (2a – 1) cm, q = 2 2a cm and = {(4a2 + 1– 4a) + 8a}cm2 D = (4a2 + 4a + 1)cm2 = (2a + 1)2 cm2 = r2.  (p2 + q2) = r2. Hence, the given triangle is right angled. C B F E Construction : Construct a triangle DEF such that DE = AB, EF = BC and E = 90º, Proof : In order to prove that B = 90º, it is sufficient to show that ABC ~ DEF. For this we proceed as follows : Ex.59 A man goes 10 m due east and then 24 m due north. Find the distance from the starting point. Sol. Let the initial position of the man be O and his final position be B. Since the man goes 10 m due east and then 24 m due north. Therefore, AOB is a right triangle right-angled at A such that OA = 10 m and AB = 24 m. N Since DEF is a right angled triangle with right angle at E. Therefore, by Pythagoras theorem, we have DF2 = DE2 + EF2  DF2 = AB2 + 24 m W E S (By construction)]  DF2 = AC2 [ AB2 + BC2 = AC2 (Given)]  DF = AC ....(i) Thus, in ABC and DEF, we have AB = DE, BC = EF [By construction] and,AC = DF O 10m A BC2 [ DE = AB and EF = BC  B [From equation (i)] By Phythagoras theorem, we have OB2 = OA2 + AB2  OB2 = 102 + 242 = 100 + 576 = 676  OB = 676 = 26 m Hence, the man is at a distance of 26 m from the starting point. Ex.60 Two towers of heights 10 m and 30 m stand on a plane ground. If the distance between their feet is 15 m, find the distance between their tops. Sol. Given : A ABC in which B is an acute angle and AD  BC. Sol. To Prove : AC2 = AB2 + BC2 – 2BC × BD. Proof : Since ADB is a right triangle rightangled at D. So, by Pythagoras theorem, we have AC2 = (15)2 + (20)2 = 625  AC = 25 m. AB2 = AD2 + BD2 A ....(i) Again ADC is a right triangle right angled at D. A 15 cm C D E 15 cm B B So, by Pythagoras theorem, we have Ex.61 In Fig., ABC is an obtuse triangle, obtuse angled at B. If AD  CB, prove that Sol. C D AC2 = AD2 + DC2 AC2 = AB2 + BC2 + 2BC × BD  AC2 = AD2 + (BC – BD)2 Given : An obtuse triangle ABC, obtuseangled at B and AD is perpendicular to CB produced.  AC2 = AD2 + (BC2 + BD2 – 2BC · BD)  AC2 = (AD2 + BD2) + BC2 – 2BC · BD  To Prove : AC2 = AB2 + BC2 + 2BC × BD Proof : Since ADB is a right triangle right angled at D. Therefore, by Pythagoras theorem, we have AB2 = AD2 + DB2 ....(i) A  AC2 = AB2 + BC2 – 2BC · BD [Using (i)] Hence, AC2 = AB2 + BC2 – 2BC · BD Ex.63 If ABC is an equilateral triangle of side a, 3 prove that its altitude = a. 2 Sol. ABD is an equilateral triangle. We are given that AB = BC = CA = a. D AD is the altitude, i.e., AD  BC. C B Again ADC is a right triangle right angled at D. Therefore, by Phythagoras theorem, we have AC2 = AD2 + DC2  = AD2 + DB2 AB = AC and AD = AD + BC2 + 2BC · BD  AC2 = AB2 + BC2 + 2BC · BD [Using (i)] Hence, AC2 = AB2 + BC2 + 2BC · BD Ex.62 In figure, B of ABC is an acute angle and AD  BC, prove that AC2 = AB2 + BC2 – 2BC × BD (Given) (Common side)  ABD ACD (By RHS congruence)  AC2 = AD2 + (DB + BC)2 AC2 Now, in right angled triangles ABD and ACD, we have  BD = CD  BD = DC = 1 a BC = 2 2 A Since the diagonals of a rhombus bisect each other at right angles. AOB = BOC = COD = DOA = 90º and AO = CO, BO = OD. Since AOB is a right triangle right-angle at O. B C D C D From right triangle ABD. a AB2 = AD2 + BD2  a2 = AD2 +   2  AD2 = a2 – 2 a2 3 3 = a2  AD = a. 4 4 2 Ex.64 ABC is a right-angled triangle, right-angled at A. A circle is inscribed in it. The lengths of the two sides containing the right angle are 5 cm and 12 cm. Find the radius of the circle. Given that ABC is right angled at A. Sol. AC = 5 cm and AB = 12 cm BC2 = AC2 + AB2 = 25 + 144 = 169  BC = 13 cm Join OA, OB, OC C O B A  AB2 = OA2 + OB2 2 2 1  1  AB2=  AC  +  BD  OA  OC  2 2     and OB  OD  4AB2 = AC2 + BD2 ....(i) Similarly, we have 4BC2 = AC2 + BD2 ....(ii) 4CD2 = AC2 + BD2 ....(iii) and,4AD2 = AC2 + BD2 ....(iv) Adding all these results, we get r 4(AB2 + BC2 + AD2) = 4 (AC2 + BD2) r AB2 + BC2 + CD2 + DA2 = AC2 + BD2 O r A B Let the radius of the inscribed circle be r Area of ABC = Area of OAB + Area of OBC + Area of OCA  1 × AB × AC 2 1 1 1 = (12 × r) + (13 × r) + (5 × r) 2 2 2  12 × 5 = r × {12 + 13 + 5}    60 = r × 30 r = 2 cm Ex.65 ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2 Sol. Let the diagonals AC and BD of rhombus ABCD intersect at O. Ex.66 P and Q are the mid-points of the sides CA and CB respectively of a ABC, right angled at C. Prove that : (i) 4AQ2 = 4AC2 + BC2 (ii) 4BP2 = 4BC2 + AC2 (iii) (4AQ2 + BP2) = 5AB2 Sol. (i) Since AQC is a right triangle right-angled at C.  AQ2 = AC2 + QC2  4AQ2 = 4AC2 + 4QC2 [Multiplying both sides by 4]  4AQ2 = 4AC2 + (2QC)2  4AQ2 = 4AC2 + BC2 [ BC = 2QC] (ii) Since BPC is a right triangle right-angled at C. A A E F P B Q B C and,OC2 = OD2 + CD2  4BP2 = 4BC2 + 4CP2 OB2 – OC2 = (OD2 + BD2) – (OD2 + CD2) [Multiplying both sides by 4]  OB2 – OC2 = BD2 – CD2  4BP2 = 4BC2 + (2CP)2 4AQ2 = 4AC2 + BC2 and, 4BC2 = 4BC2 + AC2 ....(iii)  (BD2 + CE2 + AF2) – (AE2 + CD2 + BF2) = 0 [In ABC, we have AB2 = AC2 + BC2] AF2 + BD2 + CE2 = AE2 + BF2 + CD2 Ex.68 In a right triangle ABC right-angled at C, P and Q are the points on the sides CA and CB respectively, which divide these sides in the ratio 2 : 1. Prove that (i) 9 AQ2 = 9 AC2 + 4 BC2 + OC2 – OD2 – OE2 – OF2 (ii) 9 BP2 = 9 BC2 + 4 AC2 (ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2 (i) In right triangles OFA, ODB and OEC, we have and,OA2 – OB2 = AF2 – BF2 = (BD2 – CD2) + (CE2 – AE2) + (AF2 – BF2)  4(AQ2 + BP2) = 5 AB2 Let O be a point in the interior of ABC and let OD  BC, OE  CA and OF  AB. ....(ii) (OB2 – OC2) + (OC2 – OA2) + (OA2 – OB2)  4(AQ2 + BP2) = 5 (AC2 + BC2) (i) AF2 + BD2 + CE2 = OA2 + OB2 OC2 – OA2 = CE2 – AE2 Adding (i), (ii) and (iii), we get  4AQ2 + 4BP2 = (4AC2 + BC2) + (4BC2 + AC2) Ex.67 From a point O in the interior of a ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that : [NCERT] (iii) 9 (AQ2 + BP2) = 13 AB2 Sol. It is given that P divides CA in the ratio 2 : 1. Therefore, 2 AC 3 CP = OA2 = AF2 + OF2 OB2 = BD2 + OD2 and,OC2 = CE2 + OE2 Adding all these results, we get ....(i) Also, Q divides CB in the ratio 2 : 1.  QC = 2 BC 3 ....(ii) A OA2 + OB2 + OC2 = AF2 + BD2 + CE2 + OF2 + OD2 + OE2 P  AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2 (ii) In right triangles ODB and ODC, we have ....(i) Similarity, we have [ AC = 2CP] (iii) From (i) and (ii), we have Sol. C OB2 = OD2 + BD2  BP2 = BC2 + CP2  4BP2 = 4BC2 + AC2 D B Q C Hence,  ABC, is a right triangle right-angled at A. (i) Applying pythagoras theorem in right-angled triangle ACQ, we have AQ2 = QC2 + AC2   AQ2 = 4 BC2 + AC2 [Using (ii)] 9  9 AQ2 = 4 BC2 + 9 AC2 ....(iii) Ex.70 The perpendicular AD on the base BC of a ABC intersects BC at D so that DB = 3 CD. Prove that 2AB2 = 2AC2 + BC2. Sol. We have, A (ii) Applying pythagoras theorem in right triangle BCP, we have BP2 = BC2 + CP2  BP2 = BC2 +  4 AC2 [Using (i)] 9  9 BP2 = 9 BC2 + 4 AC2 ....(iv) B Adding (iii) and (iv), we get C DB = 3CD  BC = BD + DC 9 (AQ2 + BP2) = 13 (BC2 + AC2)  BC = 3 CD + CD  9 (AQ2 + BP2) = 13 AB2 [ BC2 = AC2 + AB2] Ex.69 In a ABC, AD  BC and AD2 = BC × CD. Prove that ABC is a right triangle. Sol. D In right triangles ADB and ADC, we have AB2 = AD2 + BD2 ....(i) A  BD = 4 CD  CD =  CD = 1 BC 4 1 1 BC and BD = 3CD = BC ....(i) 4 4 Since ABD is a right triangle right-angled at D.  AB2 = AD2 + BD2 ....(ii) Similarly, ACD is a right triangle right angled at D.  AC2 = AD2 + CD2 ....(iii) Subtracting equation (iii) from equation (ii) we get B D AB2 – AC2 = BD2 – CD2 C and, AC2 = AD2 + DC2 2 ....(ii) Adding (i) and (ii), we get AB2 + AC2 = 2 AD2 × BD2 + DC2   AB2 + AC2 = 2BD × CD + BD2 + DC2 [  AD2 = BD × CD (Given)]  AB2 + AC2 = (BD + CD)2 = BC2 Thus, in  ABC, we have AB2 = AC2 + BC2 2 3  1   AB2 – AC2 =  BC  –  BC  4 4     1 3   From (i) CD  4 BC, BD  4 BC    AB2 – AC2 = 9 1 BC2 – BC2 16 16  AB2 – AC2 = 1 BC2 2  2(AB2 – AC2) = BC2  2AB2 = 2AC2 + BC2 Also, Area of ABC = Ex.71 ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB, prove that   AB2 = BC2 + AC2 1 1 1 = 2 + 2 2 a b p  c2 = a2 + b2 2 Sol.(i) Let CD  AB. Then, CD = p.  ab   ab     = a2 + b2  cp  ab  c   p  p  A c D b  B a 1 1 cp = ab  cp = ab 2 2 (ii) Since ABC is right triangle right-angled at C. (i) cp = ab (ii) 1 1 (BC × AC) = ab 2 2   a 2b2 = a2 + b2  p2  a 2  b2 1  = a 2b2 p2  1 1 1 = 2 + 2 2 a b p C  Area of ABC = 1 (Base × Height) 2  Area of ABC = 1 1 (AB × CD) = cp 2 2 1 1 1 = 2 + 2  2 b a p IMPORTANT POINTS TO BE REMEMBERED 1. Two figures having the same shape but not necessarily the same size are called similar figures. 5. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle. 2. All congruent figures are similar but the converse is not true. 6. The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. 3. Two polygones having the same number of sides are similar, if (a) Their corresponding angles are equal and (b) Their corresponding sides are proportional (i.e., in the same ratio) 4. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. 7. If a line through one vertex of a triangle divides the opposite side in the ratio of other two sides, then the line bisects the angle at the vertex. 8. The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle. 9. The line drawn from the mid-point of two sides of a triangle is parallel of another side bisects the third side. 10. The line joining the mid-points of two sides of a triangle is parallel to the third side. 11. The diagonals of a trapezium divide each other proportionally. 23. The ratio of the areas of two similar triangles is equal to the ratio of 12. If a diagonals of a quadrilateral divide each other proportionally, then it is a trapezium. (i) The squares of any two corresponding sides 13. Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally. (iii) The squares of the corresponding medians. 14. If three or more parallel lines are intersected by two transversals, then the intercepts made by them on the transversals are proportional. 24. If the areas of two similar triangles are equal, then the triangles are congruent i.e., equal and similar triangles congruent. 15. AAA similarity criterion : If in two triangles, corresponding angles are equal, then the triangles are similar. 25. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other. 16. AA Similarity criterion : If in two triangles, two angles of one triangle are respectively equal the two angles of the other triangle, then the two triangles are similar. 17. SSS Similarity criterion : If in two triangles, corresponding sides are in the same ratio, then the two triangles are similar. 18. If one angle of a triangles is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar. 19. If two triangles are equiangular, then (i) The ratio of the corresponding sides is same as the ratio of corresponding median. (ii) The ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments. (iii) The ratio of the corresponding sides is same as the ratio of the corresponding altitudes. 20. If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite side in the same ratio, then the triangles are similar. 21. If two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle, then the triangles are similar. 22. If two sides and a median bisecting the third side of a triangle are respectively proportional to the two sides and the corresponding median of another triangle, then the triangles are similar. (ii) The squares of the corresponding altitudes. (iv) The squares of the corresponding angle bisector segments. 26. Pythagoras Theorem : In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 27. Converse of Pythagoras Theorem : If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to first side is a right angle. 28. In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with the twice of the square of the median which bisects the third side. 29. Three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle. 30. Three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.