Quadrilaterals Notes

QUADRILATERALS Ex.2  IMPORTANT POINTS  A quadrilateral is a figure bounded by four line segments such that no three of them are parallel. Two sides of quadrilateral are consecutive or adjacent sides, if they have a common point (vertex). Two sides of a quadrilateral are opposite sides, if they have no common end-point (vertex). The consecutive angles of a quadrilateral are two angles which include a side in their intersection. In other words, two angles are consecutive, if they have a common arm. Two angles of a quadrilateral are said to be opposite angles if they do not have a common arm. The sum of the four angles of a quadrilateral is 360º.       EXAMPLES  Ex.1  Prove that a + b = x + y. Sol. Join BD. In ABD, we have ABD + ADB = bº ....(i) In CBD, we have CBD + CDB = aº ....(ii) Adding (i) and (ii), we get (ABD +CBD) +(ADB + CDB) = aº + bº  xº + yº = aº + bº Hence, x + y = a + b Ex.3 In a quadrilateral ABCD, AO and BO are the bisectors of A and B respectively. Prove 1 that AOB = (C + D). 2 Sol. In AOB, we have  In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 2 : 4 : 5 : 7. Find the measure of each angles of the quadrilateral. The sides BA and DC of a quadrilateral ABCD are produced as shown in fig. We have A : B : C : D = 2 : 4 : 5 : 7. So, let A = 2xº, B = 4xº, C = 5xº, D = 7xº. Sol.  + B + C + D = 360º  x + 4x + 5x + 7x = 360º    18x = 360º  x = 20º  AOB = 180º – (1 + 2) Thus, the angles are :  AOB = 180º –  1 A  1 B  A = 40º, B = (4 × 20)º = 80º,    C = (5 × 20)º = 100º and,D = (7x)º = (7 × 20)º = 140º AOB + 1 + 2 = 180º 2 2  1 1    1  2 A and 2  2 B    AOB = 180º – 1 (A + B) 2 Ex.5 In a parallelogram ABCD, prove that sum of any two consecutive angles is 180º.  AOB = 180° – 1 [360º – (C + D)] 2 Sol. Since ABCD is a parallelogram. Therefore, AD || BC. [ A + B + C + D = 360º  A + B = 360º – (C + D)]  AOB = 180º – 180º +  AOB = Ex.4 1 (C + D) 2 1 (C + D) 2 In figure bisectors of B and D of quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that Now, AD || BC and transversal AB intersects them at A and B respectively.  A + B = 180º [Sum of the interior angles on the same side of the transversal is 180º] 1 P + Q = (ABC + ADC) 2 Similarly, we can prove that B + C = 180º, C + D = 180º and D + A = 180º. Sol. In PBC, we have  P + 4 + C = 180º  P + 1 B + C = 180º 2 ...(i) In QAD, we have Q + A + 1 = 180º  Q + A + 1 D = 180º 2 ...(ii) Adding (i) and (ii), we get P + Q + A + C + 1 1 + D 2 2 = 180º + 180º P + Q + A + C + 1 1 + D = 360º 2 2  P + Q + A + C + 1 (B + D) 2 = A + B + C + D [In a quadrilateral ABCD A + B + C + D = 360º] P + Q = 1 (B + D) 2 P + Q = 1 (ABC + ADC) 2 Ex.6 In a parallelogram ABCD, D = 115º, determine the measure of A and B. Sol. Since the sum of any two consecutive angles of a parallelogram is 180º. Therefore, Ex.8 ABCD is a parallelogram and line segments AX,CY are angle bisector of A and C respectively then show AX || CY. Sol. Since opposite angles are equal in a parallelogram. Therefore, in parallelogram ABCD, we have A = C A + D = 180º and A + B = 180º Now, A + D = 180º A + 115º = 180º [D = 115º (given)]  A = 65º and A + B = 180º  65º + B = 180º B = 115º Thus, A = 65º and B = 115º Ex.7 In figure, AB = AC, EAD = CAD and CD || AB. Show that ABCD is a parallelogram.  1 1 A = C 2 2  1 = 2 ....(i) [AX and CY are bisectors of A and C respectively] Now, AB || DC and the transversal CY intersects them.  2 = 3 In ABC, AB = AC Sol.     ABC = ACB [ Alternate interior angles are equal] ….(1) From (i) and (ii), we get (Angles opposite the equal sides are equal) 1 = 3 EAD = CAD[Given] ….(2) Now, EAC = ABC + ACB  An exterior angle is equal to sum of     two interior opposite angles of a triangles    EAD + CAD = ABC + ACB  CAD + CAD = ACB + ACB By (1) and (2)   2CAD = 2ACB   CAD = ACB   BC | | AD Also, CD | | AB ....(ii) [Given] [Given] Thus, we have both pairs of opposite sides of quadrilateral ABCD parallel. Therefore, ABCD is a parallelogram. Thus, transversal AB intersects AX and YC at A and Y such that 1 = 3 i.e. corresponding angles are equal.  AX || CY Ex.9 In the adjoining figure, a point O is taken inside an equilateral quad. ABCD such that OB = OD. Show that A, O and C are in the same straight line. Sol. Given a quad. ABCD in which AB = BC = CD = DA and O is a point within it such that OB = OD. Ex.11 In figure, ABCD is a trapezium such that AB || CD and AD = BC. To prove AOB + COB = 180º Proof In OAB and OAD, we have AB = AD (given), OA = OA (common) and OB = OD (given)  OAB  OAD  AOB = AOD ….(i) (c.p.c.t.) BE || AD and BE meets BC at E. Similarly, OBC ODC COB = COD ….(ii) Show that (i) ABED is a parallelogram. Now, AOB + COB + COD + AOD = 360º Sol.  2(AOB + COB) = 360º AOB + COB = 180º Ex.10 In figure AN and CP are perpendiculars to the diagonal BD of a parallelogram ABCD. Prove that : (i) ADN CBP (ii) A + C = B + D = 180º. [ at a point] Here, AB || CD (Given)  AB || DE ….(1) Also, BE || AD (Given) ….(2) From (1) and (2), (ii) AN = CP ABED is a parallelogram Sol.  AD = BE ….(3) Also, AD = BC (Given) ….(4) From (3) and (4), Since ABCD is a parallelogram.  AD || BC BE = BC Now, AD || BC and transversal BD intersects them at B and D.  BEC = BCE  1 = 2 Also, BAD = BED [Alternate interior angles are equal] ….(5) (opposite angles of parallelogram ABED) Now, in s ADN and CBP, we have i.e., 1 = 2 AND = CPD and, AD = BC [ Opposite sides of a ||gm BED = BAD Now, BED + BEC = 180º (Linear pair of angles) are equal] So, by AAS criterion of congruence ADN CBP AN = CP [Corresponding parts of congruent triangles are equal] ….(6)  BAD + BCE = 180º By (5) and (6)  A + C = 180º Similarly, B + D = 180º Ex.12 In figure ABCD is a parallelogram and DAB = 60º. If the bisectors AP and BP of angles A and B respectively, meet at P on CD, prove that P is the mid-point of CD. Sol. Ex.13 Prove that the line segments joining the midpoint of the sides of a quadrilateral forms a parallelogram. We have, DAB = 60º A + B = 180º Sol.  60º + B = 180º B = 120º Now, AB || DC and transversal AP intersects them. Points E, F, G and H are the mid-points of the sides AB, BC, CD and DA respectively, of the quadrilateral ABCD. We have to prove that EFGH is a parallelogram.  PAB = APD  APD = 30º [ PAB = 30º] Thus, in APD, we have PAD = APD [Each equal to 30º]  AD = PD .... (i) [ Angles opposite to equal sides are equal] Since BP is the bisector of B. Therefore, Join the diagonal AC of the quadrilateral ABCD. ABP = PBC = 60º Now, AB || DC and transversal BP intersects them.  CPB = ABP  CPB = 60º [ABP = 60º] Thus, in CBP, we have CBP = CPB [Each equal to 60º] Now, in ABC, we have E and F mid-points of the sides BA and BC.  EF || AC and EF = 1 AC 2 .... (1) Similarly, from ADC, we have  CP = BC GH || AC [Sides opp, to equal angles are equal]  CP = AD .... (ii) [ ABCD is a ||gm  AD = BC] and  P is the mid point of CD. 1 AC 2 ….(2) Then from (1) and (2), we have EF || GH From (i) and (ii), we get PD = CP GH = and EF = GH This proves that EFGH is a parallelogram. Ex.14 In figure ABCD is a parallelogram and X, Y are the mid-points of sides AB and DC respectively. Show that AXCY is a parallelogram. Sol. Since X and Y are the mid-points of AB and DC respectively. Therefore, 1 1 AX = AB and CY = DC ... (i) 2 2 But,AB = DC Ex.16 In figure ABCD is a parallelogram and X and Y are points on the diagonal BD such that DX = BY. Prove that (i) AXCY is a parallelogram (ii) AX = CY, AY = CX (iii) AYB CXD Given : ABCD is a parallelogram. X and Y are points on the diagonal BD such that DX = BY Sol. [ ABCD is a ||gm ] To Prove : (i) AXCY is a parallelogram (ii) AX = CY, AY = CX 1 1  AB = DC 2 2  AX = CY .... (ii) Also, AB || DC  AX || YC .... (iii) Thus, in quadrilateral AXCY, we have AX || YC and AX = YC [From (ii) and (iii)] Hence, quadrilateral AXCY is a parallelogram. Ex.15 Prove that the line segments joining the midpoints of the sides of a rectangle forms a rhombus. Sol. P, Q, R and S are the mid-points of the sides AB, BC, CD and DA of the rectangle ABCD. (iii) AYB CXD C Construction : join AC to meet BD at O. Proof : (i) We know that the diagonals of a parallelogram bisect each other. Therefore, AC and BD bisect each other at O. C D X O Y A B  OB = OD But,BY = DX  OB – BY = OD – DX  OY = OX Thus, in quadrilateral AXCY diagonals AC and XY are such that OX = OY and OA = OC i.e. the diagonals AC and XY bisect each other. Here, AC = BD (ABC  BAD) 1 AC 2 1 and PQ || AC and PQ = AC 2 1  SR || PQ and SR = PQ = AC 2 1 Similarly, PS || QR and PS = QR = BD 2   SR || PQ, PS || QR Now, and SR || AC and SR = SR = PQ = PS = QR (AC = BD) PQRS is a rhombus. Hence, AXCY is a parallelogram. (ii) Since AXCY is a parallelogram  AX = CY and AY = CX (iii) In triangles AYB and CXD, we have AY = CX [From (ii)] AB = CD [ ABCD is a parallelogram] BY = DX [Given] So, by SSS-criterion of congruence, we have AYB CXD Ex.17 In fig. ABC is an isosceles triangle in which AB = AC. CP || AB and AP is the bisector of exterior CAD of ABC. Prove that PAC = BCA and ABCP is a parallelogram. Ex.18 In the given figure, ABCD is a square and PQR = 90º. If PB = QC = DR, prove that D R C Given : An isosceles ABC having AB = AC.AP is the bisector of ext CAD and CP || AB. Sol. D To Prove : PAC = BCA and ABCP Proof : In ABC, we have AB = AC   1 = 2 A [Given] .... (i)  Angles opposite to equal   sides in a  are equal P B (i) QB = RC, (ii) PQ = QR, (iii) QPR = 45º. Sol. BC = DC, CQ = DR  BC – CQ = CDR  QB = RC From CQR, RQB = QCR + QRC Now, in  ABC, we have  RQP + PQB = 90º + QRC ext CAD = 1 + 2  90º + PQB = 90º + QRC Now, RCQ  QBP and therefore, QR = PQ PQ = QR QPR = PRQ Bur, QPR + PRQ = 90º. So, QPR = 45º  An exteriorangles is equal to the  sum of two oppositeint erior angles   ext CAD = 22 [1 = 2 (from (i))]  23 = 22 [AP is the bisector of ext.CAD CAD = 23]   3 = 2 Thus, AC intersects lines AP and BC at A and C respectively such that 3 = 2 i.e., alternate interior angles are equal. Therefore, AP || BC. But,CP || AB [Gvien] Thus, ABCP is a quadrilateral such that AP || BC and CP || AB. Hence, ABCP is a parallelogram.  EXA MPLES  Ex.19 Prove that in a parallelogram (i) opposite sides are equal Ex.20 If the diagonals of a parallelogram are perpendicular to each other, prove that it is a rhombus. Since the diagonals of a ||gm bisect each other, Sol. (ii) opposite angles are equal C D (iii) each diagonal bisects the parallelogram Sol. Given : A ||gm ABCD in which AB || DC and AD || BC. O To prove (i) AB = CD and BC = AD; B A (ii) B = D and A = C, we have, OA = OC and OB = OD. (iii) ABC = CDA and ABD = CDB Now, in AOD and COD, we have Construction join A and C. OA = OC, AOD = COD =90° In ABC and CDA, we have, and OD is common  AOD COD  AD = CD (c.p.c.t.) Now, AB = CD and AD = BC (opp. sides of a ||gm) and AD = CD (proved)  AB = CD = AD = BC Hence, ABCD is a rhombus. 1 = 2 [Alt. int. , as AB || DC and CA cuts them] Ex.21 PQRS is a square. Determine SRP. 3 = 4 Sol. PQRS is a square. [Alt. int. , as BC || AD and CA cuts them]  PS = SR and PSR = 90° AC = CA (common) Now, in  PSR, we have PS = SR ABC CDA [AAS-criterial] (i) ABC CDA (proved)  AB = CD and BC = AD (c.p.c.t.) (ii) ABC CDA (proved) B = D (c.p.c.t.) Also, 1 = 2 and 3 = 4 1 + 4 = 2 + 3 A = C  Hence, B = D and A = C (iii) Since ABC CDA and congruent triangles are equal in area,  Angles opp. to   equal sides are equal    But,1 + 2 + PSR = 180º   21 + 90° = 180º  21 = 90° So we have ABC = CDA Similarly, ABD = CDB  1 = 2   1 = 45° [PSR = 90°] Ex.22 In the adjoining figure, ABCD is a rhombus. If A = 70º, find CDB Sol. Ex.24 Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides. Sol. We have C = A = 70º (opposite  of a ||gm) Let CDB = xº In CDB, we have CD = CB CBD = CDB = xº  CDB + CBD + DCB = 180º (angles of a triangle)  xº + xº + 70º = 180º  2x = 110, i.e., x = 55 Hence, CDB = 55º Ex.23 ABCD is a rhombus with ABC = 56°. Determine ACD. Sol. ABCD is a parallelogram C D To Prove : 1 (AB – DC) 2 Construction : Join DP and produce DP to meet AB in R. Proof : Since AB || DC and transversal AC cuts them at A and C respectively. 1 = 2 .... (i) [Alternate angles are equal] Now, in s APR and DPC, we have 1 = 2 [From (i)] (ii) PQ = AP = CP  ADC = 56° [ABC = 56° (Given)] [ODC = [ P is the mid-point of AC]  Corresponding parts of   congruent triangles are equal    A  ABC = ADC  ODC = 28° (i) PQ || AB or DC and, 3 = 4 [Vertically opposite angles] So, by ASA criterion of congruence APR DPC  AR = DC and PR = DP ....(ii) O B Given : A trapezium ABCD in which AB || DC and P and Q are the mid-points of its diagonals AC and BD respectively. 1 ADC] 2 Now, OCD we have, OCD + ODC + COD = 180°  ODC + 28° + 90° = 180°  OCD = 62° ACD = 62°.  The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.  The line drawn through the mid-point of one side of a triangle, parallel to another side, intersects the third side at its mid-point. In DRB, P and Q are the mid-points of sides DR and DB respectively.  PQ || RB  PQ || AB [ RB is a part of AB]  PQ || AB and DC [ AB || DC (Given)] This proves (i). Again, P and Q are the mid-points of sides DR and DB respectively in DRB.  PQ = 1 1 RB PQ = (AB – AR) 2 2  PQ = 1 (AB – DC) [From (ii), AR = DC] 2 This proves (ii). A  A diagonal of a parallelogram divides it into two triangles of equal area.  For each base of a parallelogram, the corresponding altitude is the line segment from a point on the base, perpendicular to the line containing the opposite side.  Parallelograms on the same base and between the same parallels are equal in area.  A parallelogram and a rectangle on the same base and between the same parallels are equal in area.  The area of a parallelogram is the product of its base and the corresponding altitude.  Parallelograms on equal bases and between the same parallels are equal in area. B  EXAMPLES  D C DE = 1 AB DE = AF = BF 2 .... (i) EF = 1 BC  EF = BD = CD 2 .... (ii) DF = 1 AC DF = AE = EC 2 ....(iii) Now, in s DEF and AFE, we have Ex.25 In the adjoining figure, ABCD is parallelogram and X, Y are the points on diagonal BD such that DX = BY. Prove that CXAY is a parallelogram. DE = AF [From (i)] DF = AE [From (ii)] and,EF = FE [Common] C D So, by SSS criterion of congruence, X  O Y A Since the diagonals of a parallelogram bisect each other, we have OA = OC and OD = OB.  DEF AFE Similarly, DEF FBD EDC B Join AC, meeting BD at O. Sol. E F and DEF Hence, AFE FBD EDC DEF Now, OD = OB and DX = BY Ex.27 In fig, AD is the median and DE || AB. Prove that BE is the median.  OD – DX = OB – BY  OX = OY Sol. Now, OA = OC and OX = OY In order to prove that BE is the median, it is sufficient to show that E is the mid-point of AC.  CXAY is a quadrilateral whose diagonals bisect each other. Now, AD is the median in ABC  CXAY is a ||gm  D is the mid-point of BC. Ex.26 Prove that the four triangles formed by joining in pairs, the mid-points of three sides of a triangle, are concurrent to each other. Given : A triangle ABC and D,E,F are the midpoints of sides BC, CA and AB respectively. Sol. To Prove :    AFE FBD EDC DEF. Proof : Since the segment joining the midpoints of the sides of a triangle is half of the third side. Therefore, Since DE is a line drawn through the midpoint of side BC of ABC and is parallel to AB (given). Therefore, E is the mid-point of AC. Hence, BE is the median of ABC. Ex.28 Let ABC be an isosceles triangle with AB = AC and let D,E,F be the mid-points of BC, CA and AB respectively. Show that AD FE and AD is bisected by FE. ABCD is a parallelogram. Sol. Given : An isosceles triangle ABC with D, E and F as the mid-points of sides BC, CA and AB respectively such that AB = AC. AD intersects FE at O. Sol.  AD = BC and AD || BC 1 1  AD = BC and AD || BC 3 3  AP = CQ and AP || CQ Thus, APCQ is a quadrilateral such that one pair of opposite side AP and CQ are parallel and equal. Hence, APCQ is a parallelogram. To Prove : AD  FE and AD is bisected by FE. Constructon : Join DE and DF. Proof : Since the segment joining the mid-points of two sides of a triangle is parallel to third side and is half of it. Therefore, DE || AB and DE = 1 AB 2 Also, DF || AC and DF = 1 AC 2 Ex.30 In fig. D,E and F are, respectively the midpoints of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle. Since the segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. Sol. A E F  But,AB = AC [Given] 1 1 AB = AC DE = DF .... (i)  2 2 1 Now, DE = AB  DE = AF .... (ii) 2 1 ....(iii) and,DF = AC DF = AE 2 From (i), (ii) and (iii) we have DE = AE = AF = DF  DEAF is a rhombus.  Diagonals AD and FE bisect each other at right angle. AD FE and AD is bisected by FE. B  DE = D C 1 AB 2 .... (i) E and F are the mid-points of AC and AB respectively.   EF = 1 BC 2 .... (ii) F and D are the mid-points AB and BC respectively.   FD = 1 AC 2 Now, ABC is an equilateral triangle Ex.29 ABCD is a parallelogram. P is a point on AD 1 such that AP = AD and Q is a point on BC 3 1 such that CQ = BP. Prove that AQCP is a 3 parallelogram.  AB = BC = CA  1 AB  1 BC  1 CA 2 2 2  DE = EF = FD [Using (i), (ii) and (iii)] Hence, DEF is an equilateral triangle. Ex.31 P,Q and R are, respectively, the mid-points of sides BC, CA and AB of a triangle ABC. PR and BQ meet at X. CR and PQ meet at Y. 1 Prove that XY = BC 4 Sol. Ex.32 Show that the quadrilateral, formed by joining the mid-points of the sides of a square, is also a square. Given : A square ABCD in which P, Q, R, S are the mid-points of sides AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined. Sol. Given : A ABC with P,Q and R as the mid-points of BC, CA and AB respectively. PR and BQ meet at X and CR and PQ meet at Y. To Prove : PQRS is a square. Construction : Join AC and BD. Construction : Join “X and Y. A R Q Y X B P Proof : In ABC, P and Q are the mid-points of sides AB and BC respectively. C Proof: Since the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of it. Therefore, Q and R are mid-points of AC and AB respectively.  RQ || BC and RQ = 1 BC 2  PQ || AC and PQ = 1 AC 2 .... (i) In ADC, R and S are the mid-points of CD and AD respectively.  RS || AC and RS = .... (i) 1 AC 2 ....(ii) From (i) and (ii), we have   P is the mid  po int    1 of BC BC  BP   2  PQ || RS and PQ = RS ....(iii)  RQ || BP and RQ = BP Thus, in quadrilateral PQRS one pair of opposite sides are equal and parallel.  BPQR is a parallelogram. Hence, PQRS is a parallelogram. Since the diagonals of a parallelogram bisect each other. Now, in s PBQ and RCQ, we have  X is the mid-point of PQ.  ABCD, is a squareAB  BC  CD  DA   1   AB  1 CD and 1 AB  1 BC  2 2 2  2  PB = RC   X is the po int of int er sec tion of    gm  diagonals BQ and PR of || BPQR  Similarly, Y is the mid-point of PQ. BQ = CQ Now, consider PQR. XY is the line segment joining the mid-points of sides PR and PQ. and PBQ = RCQ [Each equal to 90°]  XY = 1 RQ 2 1 But RQ = BC 2 Hence, XY = 1 BC. 4 So, by SAS criterion of congruence PBQ RCQ .... (i) [From (i)] [PB = CR and BQ = CQ]    PQ = QR   Corresponding parts of congruent s are equal From (iii) and (iv), we have PQ = QR = RS ....(iv) But,PQRS is a||gm. A QR = PS So, PQ = QR = RS = PS P ....(v) Now, PQ || AC [From (i)]  PM || NO C ....(vi)  AQP = ABC [Corresponding angles]   PM || MO B Proof : Since PQ || BC. Therefore, Since P and S are the mid-points of AB and AD respectively. PS || BD Q  AQP = 90° [ABC = 90°] ....(vii) But,AQP + BQP = 180° Thus, in quadrilateral PMON, we have PM || NO [From (vi)] PN || MO [From (vii)] So, PMON is a parallelogram. [ AQP & BQP are angles of a linear pair]  90° + BQP = 180°   BQP = 90°  MPN = MON Thus, AQP = BQP = 90°  MPN = BOA [MON = BOA] Now, in ABC, P is the mid-point of AC and PQ || BC. Therefore, Q is the mid-point of AB i.e, AQ = BQ.  MPN = 90° z  Diagonals of square are    AC  BD  BOA  90  Consider now s APQ and BPQ. we have, AQ = BC  [Proved above]  QPS = 90° AQP = BQP and, PQ = PQ Hence, PQRS is a square. So, by SAS cirterion of congruence Ex.33 ABC is a triangle right angled at B ; and P is the 1 mid-point of AC. Prove that PB = PA = AC. 2 Sol. [From (i)] Thus, PQRS is a quadrilateral such that PQ = QR = RS = SP and QPS = 90°. Given : ABC right angled at B, P is the midpoint of AC. To Prove : PB = PA = 1 AC. 2 Construction : Through P draw PQ || BC meeting AB at Q. APQ BPQ  PA = PB Also, PS = 1 AC, since P is the mid-point of AC 2 Hence, PA = PB = 1 AC. 2 Ex.34 Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus. Sol. PQ || SR and PQ = SR  PQRS is a parallelogram. Given : A rectangle ABCD in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.  AD = BC  To Prove : PQRS is rhombus.  AS = BQ Construction : Join AC. In s APS and BPQ , we have Proof : In ABC, P and Q are the mid-points of sides AB and BC respectively. AP = BP 1 1 AD = BC 2 2 ....(iv) [P is the mid-point of AB] .... (i) and, AS = BQ In  ADC, R and S are the mid-points of CD and AD respectively. R Now, ABCD is a rectangle. PAS = PBQ [Each equal to 90°] 1  PQ || AC and PQ = AC 2 D A APS BPQ C Q P ....(v) [ Corresponding parts of congruent triangles are equal] B 1  SR || AC and SR = AC 2 From (i) and (ii), we get [From (iv)] So, by SAS criterion of congruence PS = PQ S ....(iii) .... (ii) From (iii) and (v), we obtain that PQRS is a parallelogram such that PS = PQ i.e., two adjacent sides are equal. Hence, PQRS is a rhombus. IMPORTANT POINTS TO BE REMEMBERED 1. Sum of the angles of a quadrilateral is 360º. 2. A diagonal of a parallelogram divides it into two congruent triangles. 3. Two opposite angles of a parallelogram are equal. 4. The diagonals of a parallelogram bisect each other. 13. Each of the four sides of a rhombus of the same length. 14. The diagonals of a rectangle are of equal length. 15. Diagonals of a parallelogram are equal if and only if it is a rectangle. 5. In a parallelogram, the bisectors of any two consecutive angles intersect at right angle. 16. The diagonals of a rhombus are perpendicular to each other. 6. If a diagonal of a parallelogram bisects one of the angles of the parallelogram it also bisects the second angle. 17. Diagonals of a parallelogram are perpendicular if and only if it is a rhombus. 7. The angles bisectors of a parallelogram form a rectangle. 8. A quadrilateral is a parallelogram if its opposite sides are equal. 9. A quadrilateral is a parallelogram iff its opposite angles are equal. 10. The diagonals of a quadrilateral bisect each other, iff it is a parallelogram. 11. A quadrilateral is a parallelogram if its one pair of opposite sides are equal and parallel. 12. Each of the four angles of a rectangle is a right angle. 18. The diagonals of a square are equal and perpendicular to each other. 19. If the diagonals of a parallelogram are equal and intersect at right angle, then it is a square. 20. The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. 21. A line through the mid-point of a side of a triangle parallel to another side bisects the third side. 22. The quadrilateral formed by joining the midpoints of the sides of a quadrilateral, in order, is a parallelogram.