Question

In a $△ABC$, let $P$ and $Q$ are points on $AB$ and $AC$ respectively sucvh that $PQ|BC$. Prove that the median $AD$ bisects $PQ$.

Answer 1

Solution:-

Given that:- $△ABC$ in which P and Q are points on sides AB and AC respectively such that $PQ\parallel BC$ and AD is a median.

To prove:- AD bisects PQ, i.e. $PE=EQ$

Proof:-

In $△APE$ and $△ABD$

$\angle APE=\angle ABD\left[\text{Corresponding angles}\right]$

$\angle PAE=\angle BAD\left[\text{Common angles}\right]$

$\therefore △APE\sim △ABD\left[\text{AA similarity}\right]$

As we know that corresponding sides of similar triangles are proportional.

$\therefore \frac{AE}{AD}=\frac{PE}{BD}.....\left(i\right)$

Now in $△AQE$ and $△ACD$

$\angle AQE=\angle ACD\left[\text{Corresponding angles}\right]$

$\angle QAE=\angle CAD\left[\text{Common angles}\right]$

$\therefore △AQE\sim △ACD\left[\text{AA similarity}\right]$

Since corresponding sides of similar triangles are proportional.

$\therefore \frac{AE}{AD}=\frac{EQ}{DC}.....\left(ii\right)$

From ${eq}^n\left(i\right)\&\left(ii\right)$, we have

$\frac{PE}{BD}=\frac{EQ}{DC}.....\left(iii\right)$

$\because$ Ad is a median $\left[\text{Given}\right]$

$\therefore BD=DC.....\left(iv\right)$

From ${eq}^n\left(iii\right)\&\left(iv\right)$, we have

$\Rightarrow PE=EQ$

$\therefore$ AD bisects PQ

Hence proved that AD bisects PQ.