In a $△ A B C$ , Let P and Q be points on AB and AC respectively such that PQ || BC. Prove that median AD bisects PQ.

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Solution

Suppose the median AD intersects PQ at E.
Now, PQ || BC $[C o r r e s p o n d i n g a n g l e s]$

$\Rightarrow ∠ A P E = ∠ B$ and $∠ A Q E = ∠ C$

So, in $△^{'} s$ APE and ABD, we have

$∠ A P E = ∠ A B D$ and, $∠ P A E = ∠ B A D$ [common]

$∴ △ A P E \sim △ A B D$

$△ A P Q \sim △ A B C$

So, $\frac{P E}{B D} = \frac{P Q}{B C}$

$\frac{P E}{B D} = \frac{P Q}{2 B D}$

We get, PQ=2PE

Hence, proved.

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