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

**Ans: **

**Step 1: **Let’s start from comparing triangles △ APR and △ ABD. Here we have:

∠ APR = ∠ ABD (corresponding angles, ∵ PQ ǁ BC)

∠ PAR = ∠ BAD (Common angle)

∴ by AA similarity criterion, we get:

∴ △ APR ~ △ ABD

**∴ …. (i)**

Similarly, we compare △ ARQ and △ ADC:

∠ AQR = ∠ ACD (corresponding angles, ∵ PQ ǁ BC)

∠ QAR = ∠ CAD (Common angle)

∴ by AA similarity criterion, we get:

∴ △ AQR ~ △ ACD

**∴ …. (ii)**

**Step 3:** Next, by comparing equations (i) and (ii), we get:

…. (iii)

It is given that AD is the median in Δ ABC

∴ BD = CD

Hence, equation (iii) becomes:

or PR = QR

**Therefore, median AD is also bisecting PQ…. Hence Proved !**

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