Question

In $\Delta ABC$ lines are drawn through $A,B$ and $C$ parallel to sides $BC,CA$ and $AB$ respectively forming a $\Delta PQR$ prove that $BC=\frac{1}{2}QR$

Answer 1

Given that:- $ABC$ is a triangle, lines are drawn through $A,B$ and $C$ parallel respectively to the sides $BC,CA$ and $AB$ forming $△PQR$

To prove:- $BC=\frac{1}{2}QR$

Proof:-

In quadrilateral $AQBC$

$AQ\parallel CB$

$AC\parallel QB$

As we know that if the opposite sides of a quadrilateral are parallel, then it is a parallelogram.

$\therefore AQBC$ is a parallelogram

$\therefore BC=QA.....\left(1\right)\left[\text{Opposite sides of a parallelogram are equal}\right]$

Similarly again, in quadrilateral $ARCB$

$AR\parallel BC$

$AB\parallel RC$

$\therefore ARCB$ is a parallelogram

$\therefore BC=AR.....\left(2\right)\left[\text{Opposite sides of a parallelogram are equal}\right]$

Now from ${eq}^n\left(1\right)\&\left(2\right)$

$QA=AR=\frac{1}{2}QR.....\left(3\right)$

Now from ${eq}^n\left(1\right)\&\left(3\right)$, we have

$BC=\frac{1}{2}QR$

Hence proved.