Question

Abcd is a parallelogram ab is divided at p and cd at q so that ap:pb=3:2 and cd:qd=4:1 if pq meets ac at r prove ar=3/7 ac

Answer 1

Answer:

The proof is explained below.

Step-by-step explanation:

Given ABCD is a parallelogram AB is divided at P and CD at Q so that AP:PB=3:2 and CQ:QD=4:1 if PQ meets AC at R then we have to prove that  $AR=\frac{3}{7}AC$

In ΔAPR and ΔCQR

∠APR=∠CQR   (Vertically opposite angles)

∠PAR=∠QCR   (Alternate angles)

By AA similarity rule,  ΔAPR ∼ ΔCQR

Therefore, we can write the sides in proportional

⇒  $\frac{AP}{CQ} = \frac{PR}{QR}=\frac{AR}{CR}$

⇒   $\frac{AP}{CQ}=\frac{AR}{CR}$

It is also given that AP:PB=3:2 ∴ $AP=\frac{3}{5}AB$

⇒ $\frac{3AB}{5CQ} =\frac{AR}{CR}$

Also, CD:QD=4:1  ∴ $CQ=\frac{4}{5}CD=\frac{4}{5}AB$

⇒  $\frac{AR}{CR}=\frac{3}{4}$

⇒  $\frac{CR}{AR}=\frac{4}{3}$

⇒ $\frac{CR}{AR}+1=\frac{4}{3}+1=\frac{7}{3}\\\\\frac{CR+AR}{AR}=\frac{7}{3}\\\\\frac{AC}{AR}=\frac{7}{3}\\\\AR=\frac{3}{7}AC$

Hence Proved.