Triangles Notes

TRIANGLES A CONTENTS O  Introduction C B Circumcentre OA = OB = OC  Angle sum property for triangle (ii)  Congruent triangles  Criteria for congruent triangles  Isosceles triangle  (i) Orthocenter is the point of intersection of the altitudes. Note : (1) All four points are coincide for an equilateral triangle. (2) Orthocentre (H), Centroid (G), Circumcentre (O) are always collinear points and G divides OH in ratio 1 : 2. (ii) Circumcentre is the point of intersection of the perpendicular bisectors of the sides. Theorem : (iii) In centre is the point of intersection of the angular bisectors. Prove that sum of all three angles is 180° or 2 right angles. (iv) Centroid is the point of intersection of the medians. Given : (v) The circumcentre of a triangle is equidistant from its vertices. (vi) The in centre of a triangle is equidistant from its sides. (vii)The centroid divides a median in the ratio 2 : 1. (viii) The orthocentre of a right angled triangle lies at the vertex containing the right angle.  ABC To prove : A + B + C = 180° Construction : point A. Draw PQ || BC, passes through Proof : 1  B  alternate angles PQ  BC  and 3  C ........(i) Two triangles are congruent if and only if one of them can be made to superpose on the other, so as to cover it exactly.  PAQ is a line 1 + 2 + 3 = 180° (linear pair application) Thus, congruent triangles are exactly identical. B + 2 + C = 180° B + CAB + C = 180° = 2 right angles. Proved. Theorem : If one side of a triangle is produced then the exterior angle so formed is equal to the sum of two interior opposite angles. A Example 1 : If ABC ~  DEF then we have : A = D, B = E, C = F; and AB = DE, BC = EF and AC = DE. Example 2 If ABC ~  EDF then we have: A = E, B = D, C = F; and AB = ED, BC = DF and AC = EF. 1 Note : 2 B 3 (1) Every triangle is congruent to itself, i.e., 4 C D   ~ ABC. ABC  Means 4 = 1 + 2 Proof : 3 = 180° – (1 + 2) ….(1) (by angle sum property) and BCD is a line (2) If ABC ~  DEF then DEF ~  ABC. 3 + 4 = 180° (linear pair) or 3 = 180° – 4 …..(2) by (1) & (2) (4) 'c.p.c.t.' for 'corresponding parts of congruent triangles'. 180° – (1 + 2) = 180° – 4 1 + 2 = 4 Proved. Two geometrical figures having exactly the same shape & size are known as congruent figures. Lines, polygons, circles etc. can congruent. Note : (1) If radius of a circle is same as other circle then only both circles are congruent. (3) If ABC ~  DEF, and DEF ~  PQR, then ABC ~ PQR.  (1) SSS (Side Side Side) By SSS criteria ABC ~  EDF A = E, B = D, C = F (c.p.c.t.) (2) ASA (Angle Side Angle) A y x (2) Two line segment are congruent only when their length are equal. B 6cm C By ASA criteria ABC ~  DEF A = D, AB = DE, AC = DF (c.p.c.t.) (3) AAS (Angle Angle Side) A A x (2) SAS x B C F 3cm C B 3 cm y 4.3 cm 70° D y E By AAS, ABC ~  FDE D (3) T 40° R 40° 7 cm 7 cm C = E, AB = FD, AC = FE (c.p.c.t.) E F (4) SAS (Side Angle Side) X AAS (not RHS) (4) By SAS, ABC ~  QPR SAS (not RHS) A = Q, C = R, AC = QR (c.p.c.t.) (5) RHS (Right Hypotenuse Side) A 3 cm Theorem 1 : If two angles and the included side of one triangle are equal to two angles and the included side of other triangle, then both triangles are congruent. Proof : 5cm Given : ABC and DEF in which B C ABC = DEF, ACB = DFE and BC = EF. ~ QPR By RHS, ABC  D' A D D' A = Q, C = R, BC = PR (c.p.c.t.) B C E F To prove : ABC ~  DEF. Proof : Eg.:(1) Case I Let AC = DF. Not congruent ( SSA is not a rule) In this case, AC = DF, BC = EF and C = F. ~ DEF (SAS-criteria) ABC  Case II If possible, let AC DF. Ex.1 Then, construct D' F = AC. Join D' E. Now, in ABC and D'EF, we have AC = D'F, BC = EF and C = F. ABC ~  D'EF (SAS-criteria) Sol.  EXAMPLES  Prove that diagonal of a parallelogram divides it into two congruent triangles. Let ABCD is a parallelogram and AC is a diagonal. ABC = D'EF (c.p.c.t) But, ABC = DEF (given) D'EF = DEF. This is possible only when D and D' coincide. ABC ~  DEF. Theorem 2 : Two right-angled triangles are congruent if one side and the hypotenuse of the one are respectively equal to the corresponding side and the hypotenuse of the other. (i.e. RHS) Given : Two right-angled triangles ABC & DEF in which B =E = 90°, BC = EF and AC = DF. ~ DEF. To prove : ABC  Construction : Produce DE to G such that GE = AB. Join GF. (By SSS) : In ABC and ADC AB = CD (opp. sides of ||gm) BC = AD (opp. sides of ||gm) AC = AC (common) ~ CDA proved By SSS, ABC  {other results : 1 = 2, 3 = 4, B = D (c.p.c.t.)} (By ASA) : In ABC and ADC 1 = 2 (alternate) AC = AC (common) 3 = 4 (alternate)  By ASA, ABC ~  CDA {other results : B = D, AB = CD, BC = AD (c.p.c.t.)} (By AAS) : In ABC and ADC 1 = 2 (alternate) Proof : In ABC and GEF, we have : AB = GE (construction), BC = EF (given), B = FEG = 90°   ABC ~  GEF (SAS-criteria)   A = G and AC = GF (c.p.c.t.) Now, AC = GF and AC = DF GF = DF G = D A = D [ G = A] Now, A = D, B = E 3rd C = 3rd F. Thus, in ABC and DEF, we have: BC = EF, AC = DF and C = F. ~ DEF (SAS-criteria). ABC  3 = 4 (alternate) BC = AD (opp. sides)  ABC ~  CDA {other results : AB = CD, B = D, AC = AC (c.p.c.t.)} (By SAS) : In ABC and ADC AB = CD (opp. sides of ||gm) 1 = 2 (alternate) AC = AC (common) ~ CDA  ABC  {other results: 3 = 4, BC = AD, B = D (c.p.c.t.) We can not use ‘RHS’ for this proof. Note : ASS or SSA criteria for congruency is not valid. Ex.2 In Fig. it is given that AB = CF, EF = BD and ~ CBD. AFE = DBC. Prove that AFE  Ex.4 In Fig. PQRS is a quadrilateral and T and U are respectively points on PS and RS such PQ = RQ, PQT = RQU and TQS = UQS. Prove that QT = QU. Sol. We have, D A C F B E Sol. We have, AB = CF  AB + BF = CF + BF   AF = CB .... (i) In s AFE and CBD, we have  AF = CB [From (i)] AFE = DBC [Given] and EF = BD [Given] and, TQS = UQS  PQT + TQS = RQU + UQS  PQS = RQS .... (i) Thus, in triangles PQS and RQS, we have PQ = RQ So, by SAS criterion of congruence, we have [Given]  PQS = RQS [From (i)] AFE ~  CBD Ex.3 PQT = RQU In Fig. X and Y are two points on equal sides AB and AC of a ABC such that AX = AY. Prove that XC = YB. and, QS = QS [Common side] Therefore, by SAS congruence criterion, we have  PQS ~  RQS  QPS = QRS (c.p.c.t.)  QPT = QRU Sol. In s AXC and AYB, we have AX = AY [Given] A = A [Common angle] AC = AB [Given] So, by SAS criterion of congruene  AYB AXC ~  XC = YB (c.p.c.t.) ....(ii) Now, consider triangles QPT and QRS. In these two triangles, we have QP = QR [Given]  PQT = RQU [Given]  QPT = QRU [From (ii)] Therefore, by ASA congruence criterion, we get  QRU QPT ~  QT = QU. Ex.5 In Fig. PS = QR and SPQ = RQP. S [ Angles opp. to equal sides are equal] Adding (i) and (ii), we get P ACB + ACD = ABC + ADC  BCD = ABC + BDC Q [ ADC = BDC, ABC = DBC] R  BCD + BCD = DBC + BCD + BDC Prove that PR = QS and QPR = PQS. Sol. Adding BCD  on both side    In SPQ and RQP, we have PS = QR [Given] SPQ = RQP [Given]  2 BCD = 180º [Common] [ Sum of the angles of a  is 180º] PQ = PQ Therefore, by SAS criterion of congruence, we have Hence, BCD is a right angle. Ex.7 SPQ ~  RQP  SQ = RP and In Fig. AC = BC, DCA = ECB and DBC = EAC. QPR = PQS Ex.6 ABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Prove that BCD is a right angle. Sol. Given : A ABC such that AB = AC. Side BA is produced to D such that AB = AD. D Prove that triangles DBC and EAC are congruent, and hence DC = EC. A Sol. We have, DCA = ECB  DCA + ECD = ECB + ECD C B Construction : Join CD.  ECA = DCB To prove : BCD = 90º Now, in s DBC and EAC, we have Proof : In ABC, we have AB = AC ACB = ABC ...(i)   Angles opp. to  equal sides are equal   DCB = ECA BC = AC and DBC = EAC .... (i) [From (i)] [Given] [Given] Now, AB = AD [Given] So, by ASA criterion of congruence  AD = AC [ AB = AC] DBC ~  EAS Thus, in ADC, we have  DC = EC AD = AC   ACD = ADC …(ii) (c.p.c.t.) Ex.8 Sol. If the altitudes from two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.  OB = OC Given : A ABC in which altitudes BE and CF from B and C respectively on AC and AB are equal. A Now, in ABO and ACO, we have ....(ii) [ Sides opp. to equal s are equal] AB = AC [Given] OBC = OCB [From (i)] OB = OC [From (ii)] E F So, by SAS criterion of congruence B ABO ~  ACO C To prove : ABC is isoceles i.e. AB = AC  BAO = CAO [ Corresponding parts Proof : In s ABC and ACF, we have AEB = AFC BAE = CAF and, BE = CF of congruent triangles are equal] [Each equal to 90º] [Common angle] [Given] So, by AAS criterion of congurence, we have  AO is the bisector of BAC. Ex.10 In Fig. BM and DN are both perpendiculars to the segments AC and BM = DN. ABE ~  ACF   Corresponding parts of   AB = AC   congruent triangles are equal Ex.9 Sol. Hence, ABC is isosceles. In ABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that BO = CO and the ray AO is the bisector of angle BAC. In ABC, we have AB = AC Prove that AC bisects BD. In s BMR and DNR, we have Sol.  BMR = DNR A [Each equal to 90º BM AC and DN AC]  BRM = DRN [Vert. opp. angles] O and, BM = DN  B = C    [Given] C B So, by AAS criterion of congruence  Angles opposite to  equal sides are equal    1 1 B = BC 2 2  OBC = OCB  DNR BMR ~  .... (i)   OB and OC are bi sec tors of s B and C    1 1  respectively  OBC  B & OCB  C 2 2     Corresponding parts of   BR = DR   congruent triangles are equal  R is the mid-point of BD. Hence, AC bisects BD. Ex.11 In Fig. BD and CE are two altitudes of a ABC such that BD = CE. A D E Prove that ABC is isolceles. In ABD and ACE, we have ADB = AEC = 90º [Given] BAD = CAE [Common] and, BD = CE [Given] So, by AAS congruence criterion, we have ABD ~  ACE Sol. AB = AC   Corresponding parts of     congruent triangles are equal Hence, ABC is isosceles. Ex.12 If two isosceles triangles have a common base, the line joining their vertices bisects them at right angles. Sol. Given : Two isosceles triangles ABC and DBC having the common base BC such that AB = AC and DB = DC. To prove : AD (or AD produced) bisects BC at right angle. A A 1 2 3 4 E 3 = 4 3 + 4 = 180º [ Sum of the angles of a linear pair is 180º]  2 3 = 180º [ 3 = 4]  3 = 90º  3 = 4 = 90º Hence, AD bisects BC at right angles. Ex.13 AD, BE and CF, the altitudes of ABC are equal. Prove that ABC is an equilateral triangle Sol. In right triangles BCE and BFC, we have Hyp. BC = Hyp. BC BE = CF [Given] So, by RHS criterion of congruence,  BFC. BCE ~ C B 3 4 E C D Proof : In s ABD and ACD, we have AB = AC [Given] BD = CD [Given] AD = AD [Common side] So, by SSS criterion of congruence ~ ACD  ABD   1 = 2 and, But, 1 2 D B   Corresponding parts of   congruent triangles are equal BE = CE  C B Now, in s ABE and ACE, we have AB = AC [Given] 1 = 2 [From (i)] and, AE = AE [Commoni side] So, by SAS criterion of congruence,  ACE ABE ~ .... (i)   Corresponding parts of  congruent triangles are equal     Corresponding parts of   B = C   congruent triangles are equal  AC = AB .... (i) [ Sides opposite to equal angles are equal]  ABE Similarly, ABD ~ B=A [Corresponding parts of congruent triangles are equal]  AC = BC ....(ii) [Sides opposite to equal angles are equal] From (i) and (ii), we get AB = BC = AC Hence, ABC is an equilateral triangle. Ex.14 In Fig. AD = BC and BD = CA. Ex.16 In quadrilateral ABCD, AC = AD and AB bisects A. Show that ABC ~  ABD. What can you say about BC and BD ? [NCERT] Prove that ADB = BCA and DAB = CBA. Sol. In triangles ABD and ABC, we have and AD = BC [Given] BD = CA [Given] AB = AB [Common] Sol. AB = AB (common) So, by SSS congruence criterion, we have  CBA  DAB = ABC ABD ~ 1 = 2 { AB is bisector of A}   corresponding parts of    congruent triangles are equal AC = AD (Given) ~ ABD Proved By SAS, ABC   DAB = CBA also BC = BD (c.p.c.t.) Ex.15 Line-segment AB is parallel to another linesegment CD. O is the mid-point of AD (see figure). Show that (i) AOB ~  DOC (ii) O is also the mid point of BC. Sol. In ABC & ABD Ex.17 AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB. [NCERT] (i) Consider AOB and DOC ABO = DCO (Alternate angles as AB || CD and BC is the transversal) AOB = DOC (Vertically opposite angles) OA = OD (Given) Therefore, AOB ~  DOC (AAS rule) C Sol. To show CD bisect AB i.e. AO = OB in OAD and OBC O = O (vertically opposite angles) D A = B = 90° (Given) O B A (ii) OB = OC AD = BC (Given) (c.p.c.t.) So, O is the mid-point of BC. By AAS, OAD ~  OBC OA = OB (c.p.c.t.) CD, bisects AB. Proved D Ex.18 Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see figure). Show that : [NCERT] A 1 2 M l Q B A Sol. B P C (i) AMC ~  BMD (i) APB ~  AQB (ii) BP = BQ or B is equidistant from the arms of A. (ii) DBC is a right angle (iii) DBC ~  ACB (i) In APB and AQB (iv) CM = P = Q = 90° (Given) PAB = QAB (Given that ‘l’ bisect A) Sol. (i) In AMC and BMD AM = MB (M is mid point of AB) AB = AB (Common) 1 = 2 (vertically opposite angles) CM = MD (given) By SAS, AMC ~  MBD Proved. By AAS, APB ~  AQB. Proved (ii) BP = BQ (c.p.c.t.) Proved. Ex.19 In given figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. 1 AB 2 (ii) ACM = MDB (c.p.c.t. of (i)) These are alternate angles [NCERT] DB || AC E So DBC + ACB = 180° (Cointerior angles) A 2 1 DBC + 90° = 180°  DBC = 90° Proved. B Sol. D C In ABC and ADE (iii) AB = AD (Given) Given    1  2 BAC = DAE   1 DAC 2       DAC  AC = AE (Given) By SAS, ABC ~  ADE BC = DE (c.p.c.t.) Proved. Ex.20 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that: [NCERT] In DBC & ACB BC = BC (common) DBC = ACB = 90° DB = AC (c.p.c.t. of part (i)) By SAS, DBC ~  ACB. Proved (iv) DC = AB (c.p.c.t. of part (iii)) But CM = 1 DC (given) 2 CM = 1 AB Proved. 2 Construction : Draw AD, bisector of A 1 = 2 A triangle in which two sides are equal & opposite angles of these two lines are also equal. Proof : In ADB & ADC AD = AD (Common) 1 = 2 (by construction) AB = AC ~ ADC By SAS, ADB  B = C (c.p.c.t.) Proved. Note : Other result : ADB = ADC (c.p.c.t.) AB = AC = 6 cm, B = C = 70° Ex.21 Find BAC of an isosceles triangle in which 1 AB = AC and B = of right angle. 3 1 Sol. B = C = (90) = 30° 3 A + B + C = 180° (.p.) A + 30° + 30° = 180° A = 120°. Ex.22 In isosceles triangle DEF, DE = EF and E = 70° then find other two angles. D But ADB + ADC = 180° (linear pair) ADB = ADC = 90° AD BC and BD = DC (c.p.c.t.) AD is median we can say AD is perpendicular bisector of BC or we can say in isosceles , median is angle bisector and perpendicular to base also. Ex.23 ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see fig.). If AD is extended to intersect BC at P. Show that [NCERT] A Sol. D F E B Let D = F = x D + E + F = 180° (angle sum property)  x + 70° + x = 180°  2x = 110°  x = 55°  D = F = 55°. Theorem (2) : Angles opposite to equal sides of an isosceles triangle are equal. A 12 Sol. C P (i) ABD ~  ACD ~ ACP (ii) ABP  (iii) AP bisects A as well as D (iv) AP is the perpendicular bisector of BC. (i) In ABD & ACD AB = AC (ABC is isosceles  AD = AD (Common) BD = DC (DBC is isosceles ) By SSS, ABD ~  ACD Proved. (ii) In ABP & ACP AB = AC (ABC is isosceles  B D C Given : In ABC, AB = AC To prove : B = C ABP = ACP {ABC is isosceles } AP = AP (common) ~ ACP Proved. By SAS, ABP  (iii)  BAP = CAP (c.p.c.t. of part (ii)) A is bisected by AP and ADB = ADC (c.p.c.t. of part (ii)) (1) If two sides of a triangle are unequal, then the longer side has the greater angle opposite to it. CD is bisected by AP. (iv) APB = APC (c.p.c.t. of part (ii)) (2) In a triangle, the greater angle has the longer side opposite to it. but APB + APC = 180° (linear pair) APB + APB = 180° (3) Of all the line segments that can be drawn to a given line, from a point not lying on it, the perpendicular line segment is the shortest. 2APB = 180° APB = 90° = APC (4) The sum of any two sides of a triangle is greater than its third side. also PB = PC (c.p.c.t. of part (ii)) AP is perpendicular bisector of BC. (5) The difference between any two sides of a triangle is less than its third side. Proved. Ex.24 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR (see figure ). Show that: (6) Exterior angle is greater than one opposite interior angle. Ex.25 Find the relation between angles in figure. x 7 cm 5 cm y (i) ABM ~  PQN Sol. (ii) ABC ~  PQR Sol. (i) In ABM & PQN 8 cm z  yz > xz > xy x > y > z. AB = PQ (given) ( Angle opposite to longer side is greater) AM = PN (given) BM = QN ( BC = QR  Ex.26 Find the relation between the sides of triangle in figure . BC QR  ) 2 2 D 80° By SSS, ABM ~  PQN Proved. (ii) In ABC & PQR AB = PQ (given) E Sol. 70° 30° F  D > E > F B = Q (c.p.c.t. of part (i)) BC = QR (given) EF > DF > DE By SAS, ABC ~  PQR Proved. { side opposite to greater angle is longer} Ex.27 Find ACD then what is the relation between (i) ACD, ABC (ii) ACD & A ACD + 40° = 180° (linear pair) ACD = 140° Ans. also A + B = ACD (exterior angle = sum of opp. interior angles) A + 70° = 140° A = 140° – 70° A = 70° Now ACD > B Ans. ACD > A Ans. Ex.28 In Fig. E > A and C > D. Sol. Draw diagonal AC. Sol. In ABC, AB < BC { AB is smallest} 3 < 1 ……(1) {angle opp. to longer side is larger} Also in ADC AD < CD Prove that AD > EC. Sol. In ABE, it is given that E > A  AB > BE .... (i) In BCD, it is given that C > D  BD > BC ....(ii) Adding (i) and (ii), we get AB + BD > BE + BC AD > EC Ex.29 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that A > C. [NCERT] 4 < 2  CD is longest …..(2) adding equation (1) & (2) 3 + 4 < 1 + 2 C < A or A > C Proved. Ex.30 In given figure, PR > PQ and PS bisects QPR. Prove that PSR > PSQ. [NCERT] Sol. In PQR, PR > PQ B=A Q > R {angle opposite to longer side is greater} [Corresponding parts of congruent triangles are equal] and 1 = 2  AC = BC ……(1) ( PS is bisector) …..(2) ....(ii) [Sides opposite to equal angles are equal] P From (i) and (ii), we get 12 AB = BC = AC Hence, ABC is an equilateral triangle. Ex.32 In Fig. AD = BC and BD = CA. R S Q Now for PQS, PSR = Q + 1 …..(3) {exterior angle = sum of opposite interior angle} Prove that ADB = BCA and DAB = CBA. & for PSR, PSQ = R + 2 …..(4) By equation (1), (2), (3), (4), PSR > PSQ Proved. Ex.31 AD, BE and CF, the altitudes of ABC are equal. Prove that ABC is an equilateral triangle Sol. In right triangles BCE and BFC, we have Hyp. BC = Hyp. BC BE = CF [Given] So, by RHS criterion of congruence, Sol. In triangles ABD and ABC, we have and AD = BC [Given] BD = CA [Given] AB = AB [Common] So, by SSS congruence criterion, we have ABD ~  CBA  DAB = ABC   corresponding parts of  congruent triangles are equal   A  DAB = CBA F B E D Ex.33 In Fig. PQ > PR. QS and RS are the bisectors of Q and R respectively. C P  BFC. BCE ~   Corresponding parts of   B = C   congruent triangles are equal  AC = AB .... (i) [ Sides opposite to equal angles are equal] Similarly, ABD ~  ABE S Q Prove that SQ > SR. R Sol. In PQR, we have PQ > PR   ADB > ACB [Given] [ ACD = ACB] Angle opp. to larger side   of a triangle is greater  PRQ > PQR  From (i) and (ii), we get 1 1 PRQ > PQR 2 2  SRQ > SQR ADB > ABC   ADB > ABD[ ABC = ABD]  RS and QS are bi sec tors of   PRQ are PQR respectively     AB > AD [ Side opp. to greater angle is larger]  SQ > SR [ Side opp. to greater angle is larger] Ex.34 In Fig. ... (ii) [NCERT] Ex.36 Prove that any two sides of a triangle are together greater than twice the median drawn to the third side. Given : A ABC in which AD is a median. Sol. A Sol. if x > y, show that M > N. We have, LMN + xº = 180º .... (i) [Angles of a linear pair]  LNM + yº = 180º ....(ii) [Angles of a linear pair]  LMN + xº = LNM + yº But x > y. Therefore, LMN < LNM  LNM > LMN LM > LN [ Side opp. to greater angle is larger] D C B E To prove : AB + AC > 2 AD Construction : Produce AD to E such that AD = DE. Join EC. Proof : In s ADB and EDC, we have AD = DE Ex.35 In Fig. AB > AC. Show that AB > AD. [By construction] BD = DC [ D is the mid point of BC] and, ADB = EDC [Vertically opp. angles] So, by SAS criterion of congruence Sol. In ABC, we have AB > AC  ACB > ABC ADB ~  EDC [Given] .... (i) [ Angle opp. to larger side is greater] Now, in ACD, CD is produced to B, forming an ext ADB.  ADB > ACD  Exterior angle of Δ is greater     than each of interior opp. angle   Corresponding parts of  congruent triangles are equal    AB = EC Now in AEC, we have AC + EC > AE [ Sum of any two sides of a      is greater than the third]  AC + AB > 2 AD  AD  DE  AE  AD  DE  2AD and EC  AB Ex.37 In Fig. PQR is a triangle and S is any point in its interior, show that SQ + SR < PQ + PR. Sol. Ex.39 In Fig. T is a point on side QR of PQR and S is a point such that RT = ST. Given : S is any point in the interior of PQR. To Prove : SQ + SR < PQ + PR Construction : Produce QS to meet PR in T. Proof : In PQT, we have  Sum of the two sides of a Δ PQ + PT > QT   is greater than the third side   PQ + PT > QS + ST Prove that PQ + PR > QS. Sol. PQ + PR > QR .... (i)  PQ + PR > QT + RT [ QR = QT + RT] [ QT = QS + ST] In RST, we have ST + TR > SR ....(ii) Adding (i) and (ii), we get PQ + PT + ST + TR > SQ + ST + SR  PQ + (PT + TR) > SQ + SR  PQ + PR > SQ + SR SQ + SR < PQ + PR. Ex.38 In PQR S is any point on the side QR. Show that PQ + QR + RP > 2 PS. In PQR, we have  PQ + PR > QT + ST [ RT = ST (Given)] In QST, we have QT + ST > QS From (i) and (ii), we get PQ + PR > QS. Ex.40 Find OBA in given figure Sol. Sol. In PQS, we have PQ + QS > PS ... (i) [ Sum of the two sides of a  is greater than the third side] Similarly, in PRS, we have RP + RS > PS ....(ii) Adding (i) and (ii), we get (PQ + QS) + (RP + RS) > PS + PS  PQ + (QS + RS) + RP > 2 PS  PQ + QR + RP > 2 PS [ QS + RS = QR] .... (i)  AOB + 198° = 360° AOB = 360° – 198° = 162° and OA = OB = radius of circle A = B = x (let) x + x + 162° = 180° (a.s.p.) 2x + 18° x = 9°  OBA = 9°. ....(ii) IMPORTANT POINTS TO BE REMEMBERED 1. A palne figure bounded by three lines in a plane is called a triangle. 2. A triangle, no two of whose sides are equal is called a scalene triangle. 3. Atriangle whose two sides are equal is called an isosceles triangle. 4. A triangle whose sides are equal is also called an equilateral triangle. 5. A triangle with one angle a right angle is called a right angled triangle. 6. The sum of the three angles of a triangle is 180º. 7. If a side of a triangles is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles. 8. If two triangles ABC and DEF are congruent under the correspondence A D, B  E and C  F, then we write  ABC ~  DEF or ABC DEF. 9. Two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle (SAS congruence criterion). 10. Two triangles are congruent if two angles and the included side of one tringle are equal to the corresponding two angles and the included side of the other triangle (ASA congruence criterion). 11. If any two angles and non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the triangles are congruent (AAS congruence criterion). 12. If three sides of one triangle are equal to three of the other triangle, then the two triangles are congruent (SSS congruence criterion). 13. If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent (RHS congruence criterion). 14. Angles opposite to equal sides of a triangle are equal. 15. If the altitude from one vertex of a triangle bisects the opposite sides, then the triangle is isosceles. 16. In an isosceles triangle altitude from the vertex bisects the base. 17. If the bisector of the vertical angle of a triangle bisects the opposite side, then the triangle is isosceles. 18. If the altitudes of a triangles are equal, then it is equilateral. 19. In a triangle, side opposite to the larger angle is longer. 20. Sum of any two sides of a triangle is greater than the third side.