Question

In a ΔABC, P and Q are points on AB and AC respectively such that PQ || BC. Prove that the median AD, drawn from A to BC, bisects PQ.

Answer 1

Given: In the ΔABC, P and Q are points on AB and AC respectively and PQ || ВС.

To prove: E is the mid-point of PQ i.e., PE = EQ

Proof: In ΔΑΡΕ and ΔΑBD,

APE = ABD   ...(Corresponding angle)

PAE = BAD   ...(Common angle)

∴ ΔAPE ∼ ΔABD   ...(By ΑA similarity)

or `(PE)/(BD) = (AE)/(AD)`  ...(1) (By СРСТ)

Similarly, ΔAQE ∼ ΔACD

or `(QE)/(CD) = (AE)/(AD)`  ...(2) (By СРСТ) 

From (1) and (2), we get

`(PE)/(BD) = (QE)/(CD)`

or `(PE)/(BD) = (QE)/(BD)`   ...(∵ CD = BD)

or PE = QE

Hence, AD bisects PQ.

Hence Proved.