Question

In a $△ A BC$ , 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.

Approach

Since $PQ ∥ BC$ and $PR ∥ B D$ , by Basic Proportionality Theorem, $\frac{A R}{A D} = \frac{PR}{B D}$ and $\frac{A R}{A D} = \frac{RQ}{D C}$ . Since AD is median, $B D = D C$ , so $PR = RQ$ . Hence AD bisects PQ.

Step-by-step working

1
Draw correct figure with median AD and $PQ ∥ BC$ .
0.5 marks
2
Since $PQ ∥ BC$ , apply BPT in $△ A BC$ .
1 mark
3
$\frac{A R}{A D} = \frac{PR}{B D}$ and $\frac{A R}{A D} = \frac{RQ}{D C}$ .
2 marks
4
$B D = D C$ (AD is median), so $PR = RQ$ .
1 mark
5
AD bisects PQ.
0.5 marks

Concepts used

medianparallel linestrianglebasic proportionality theorem

Approach

Step-by-step working

Concepts used

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