Question

In a $△ABC$, Let P and Q be points on AB and AC respectively such that PQ || BC. Prove that median AD bisects PQ [3 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 1 Mark



The median AD intersects PQ at E.

Now, PQ || BC [Given]

$\Rightarrow \angle APE=\angle B$ and $\angle AQE=\angle C$ [Corresponding angles]

So, in $\Delta APEand\Delta ABD$,

$\angle APE=\angle ABD$ [Corresponding angles]

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

$\therefore \Delta APE\sim \Delta ABD$ [A. A Similarity]

$\Rightarrow \frac{PE}{BD}=\frac{AP}{AB}.....\left(1\right)$

Similarly,

$\Delta APQ\sim \Delta ABC$

$\Rightarrow \frac{PQ}{BC}=\frac{AP}{AB}...\left(2\right)$

From (1) and (2)

$\frac{PE}{BD}=\frac{PQ}{BC}$

$\Rightarrow \frac{PE}{BD}=\frac{PQ}{2BD}$ [$\because$ Median AD divides BC]

$\Rightarrow PQ=2PE$

$\Rightarrow$ E is the midpoint of PQ

Hence proved that median AD bisects PQ.