Question

Abcd is a parallelogram.Ab is divided at p and cd at q so that ap:pb=3:2 a d cq:qd=4:1. If pq meets ac at r, then prove that ar=3/7 ac

Answer 1

Answer:

Proved

Step-by-step explanation:

Answer 2

Opposite Sides of Parallelogram are Equal

Step-by-step explanation:

Since ABCD is a ||gm , then AB || CD and AD || BC.

Since AB || CD and AC is a transversal, so

• ∠PAR = ∠QCR     (Alternate interior angles)    .........(1)

Since ABCD is a ||gm , then AB = CD and AD = BC , as opposite sides of parallelogram are equal.

Let AB = CD = x

AP = $\frac {3x}{5}$ and PB = $\frac {2x}{5}$ (As AP:PB = 3:2)

and  

CQ = $\frac {4x}{5}$ and QD = $\frac {x}{5}$ (As CQ:QD = 4:1)

In △CQR and △APR, ∠QCR = ∠PAR        [using (1)]

∠QRC = ∠PRA       [vertically opposite angles]

⇒ △CQR ~ △APR [AA similarity]

$\frac {CR}{AR} = \frac {CQ}{AP} = \frac {QR}{PR}$ (corresponding sides of similar triangle's are proportional)

$\frac {CR}{AR} = \frac {CQ}{AP}$

$\frac {CR}{AR} = \frac {\frac {4x}{5}}{\frac {3x}{5}}$

$\frac {CR}{AR} = \frac {4}{3}$

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

$\frac {CR+AR}{AR} = \frac {7}{3}$

$7AR = 3AC$

$AR = \frac {3}{7} AC$ Proved