InterviewSolution
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InterviewSolution
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30, ABCD is a parallelogram. AB is divideP and CD at Q so that AP: PB 3:2 and CQ4:1. If PQ meets AC at R, then prove37 |
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Answer» GIVEN: ABCD is a ||gm, in which Pis a point on AB such that AP : PB = 3 : 2 and Qon CD such that CQ : QD= 4 : 1.TO PROVE:AR=3/7AC PROOF : Since ABCD is a ||gm, then AB || CD and AD || BC.Since AB || CD and AC is a transversal, so∠PAR=∠QCR(Alternateinteriorangles).........(1) Since ABCD is a ||gm, then AB = CD and AD = BC , as opposite sides of ||gmare equal.Let AB = CD =xso,AP=3x/5andPB=2x/5[as,AP:PB=3:2]andCQ=4x/5andQD=x/5[as,CQ:QD=4:1] In△CQRand△APR,∠QCR=∠PAR[using(1)]∠QRC=∠PRA[verticallyoppositeangles]⇒△CQR~△APR[AAsimilarity] ⇒CR/AR=CQ/AP=QR/PR(correspondingsidesofsimilar△'sareproportional) ⇒CR/AR=CQ/AP⇒CR/AR=4x/5/3x/5⇒CR/AR=4/3⇒CR/AR+1=4/3+1⇒CR+AR/AR=7/3⇒AC/AR=7/3⇒7AR=3AC⇒AR=3/7AC |
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