Question

In a $△ABC$, let p and q be points on AB and AC respectively such that PQ||BC. Prove that the median AD bisects PQ.

Answer 1

Suppose the median AD intersects PQ at E.

Now, $PQ||BC$

$\Rightarrow$ $\angle APE=\angle B$ and $\angle AQE=\angle C$

So, in ${△}^{\prime }sAPE$ and $ABD$, we have

$\angle APE=\angle ABD$ and,

$\angle PAE=\angle BAD$ [Common]

$\therefore$ $△APE\sim △ABD$

$\Rightarrow$ $\frac{PE}{BD}=\frac{AE}{AD}$.........(i)

Similarly, we have

$△AQE\sim △ACD$

$\therefore$ $\frac{QE}{CD}=\frac{AE}{AD}$.........(ii)

From (i) and (ii), we get

$\frac{PE}{BD}=\frac{QE}{CD}$

$\Rightarrow$ $\frac{PE}{BD}=\frac{QE}{BD}$ [$\because$ AD is the median $\therefore$ BD=CD]

$\Rightarrow$ $PE=QE$

Hence, AD bisects PQ.