Question

In $△ABC,P$ and $Q$ be a point on $AB$ and $AC$ respectively such that $PQ\parallel BC$, prove that median $AD$ bisects $PQ$.

Answer 1

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

To prove: $AD$ bisects $PQ$

Proof: In $△APE$ and $△ABD$

$\angle APE=\angle ABD$ (corresponding angles)

and $\angle PAE=\angle BAD$ (common)

$\therefore$ by using $AA$ similarity condition,

$△APE\sim △ABD$

$\Rightarrow \frac{AE}{AD}=\frac{PE}{BD}$ .......$\left(1\right)$

Now, in $△AQE$ and $△ACD$

$△AQE=△ACD$(corresponding angles)

and $\angle QAE=\angle CAD$ (common)

$\therefore$ by using $AA$ similarity condition,

$△AQE\sim △ACD$

$\Rightarrow \frac{AE}{AD}=\frac{EQ}{DC}$ .......$\left(2\right)$

Comparinr equations $\left(1\right)$ and $\left(2\right)$, we have

$\frac{PE}{BD}=\frac{EQ}{DC}$

But $BD=DC$ since $AD$ is the median.

$\Rightarrow PE=EQ$

Hence $AD$ bisects $PQ$