In triangle ABC, P and Q are points on the sides AB and AC respectively such that PQ is parallel to BC. Prove that medium AD, drawn from A to BC, bisects PQ.

Given - triangle ABC

PQ is parallel to BC

AD is median of BC ----> BD = CD

To prove - AD bisects PQ

Let us assume that the opint of intersection of AD and PQ is O

-----> we need to prove PO = QO

In triangle APQ and triangle ABC

∠PAQ = ∠ BAC (∠A is common)

∠APQ = ∠ABC (corresponding angles)

therefore, by AAA similarity criteria,

triangle APQ ~ triangle ABC

----> PQ is similar to BC BC (sides of similar triangles are proportional)

-----> if median AD bisects BC, it should bisect PQ too

-----> PO = QO

hope this is the correct answer.... plz comment if not!!!!

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