Through A, B and C lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a $Δ A B C$ as shown in figure. Show that $B C = \frac{1}{2} Q R$ .

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Solution

Given in $Δ A B C$ , PQ ∥ AB and PR ∥ AC and RQ ∥ BC

To show $B C = \frac{1}{2} Q R$ .

Proof in Quadrilateral BCAR, BR ∥CA and BC ∥ RA

So, quadrilateral , BCAR is a parallelogram.

$∴ B C = A R$ ….(i)

Now, in quadrilateral BCQA, BC ∥AQ

And AB ∥ QC

So, quadrilateral BCQA is a parallelogram.

BC = AQ …..(ii)

On adding Eqs (i) and (ii) we get

2BC = AR + AQ

$\Rightarrow 2 B C = R Q$

$\Rightarrow B C = \frac{1}{2} Q R$ . Hence proved

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