In 𝛥ABC, P and Q are points on AB and AC respectively such that

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

∴ $\frac{PR}{BD} = \frac{AR}{AD}$ …. (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

∴ $\frac{QR}{CD} = \frac{AR}{AD}$ …. (ii)

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

$\frac{PR}{BD} = \frac{QR}{CD}$ …. (iii)

It is given that AD is the median in Δ ABC

∴ BD = CD

Hence, equation (iii) becomes:

$\frac{PR}{BD} = \frac{QR}{BD}$

or PR = QR

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

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