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In a $\Delta ABC$, P and Q are points on AB and AC respectively such that $PQ \parallel BC$. Prove that the median AD, drawn from A to BC, bisects PQ.
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Correct figure [$\frac{1}{2}$ mark]
Given $PQ \parallel BC \Rightarrow PR \parallel BD$
Therefore in $\Delta APR$ and $\Delta ABD$, $\angle PAR = \angle BAD$ and $\angle APR = \angle ABD$ [$1$ mark]
$\therefore \Delta APR \sim \Delta ADB$ (By AA similarity criterion)
$\Rightarrow \frac{AR}{AD} = \frac{PR}{BD}$ (i) [$\frac{1}{2}$ mark]
Similarly $\Delta ARQ \sim \Delta ADC \Rightarrow \frac{AR}{AD} = \frac{RQ}{DC}$ (ii) [$1$ mark]
Using (i) and (ii) $\frac{PR}{BD} = \frac{RQ}{DC}$ [$\frac{1}{2}$ mark]
AD is the median $\therefore BD = DC \Rightarrow PR = RQ$ [$1$ mark]
i.e. AD bisects PQ [$\frac{1}{2}$ mark]
Given $PQ \parallel BC \Rightarrow PR \parallel BD$
Therefore in $\Delta APR$ and $\Delta ABD$, $\angle PAR = \angle BAD$ and $\angle APR = \angle ABD$ [$1$ mark]
$\therefore \Delta APR \sim \Delta ADB$ (By AA similarity criterion)
$\Rightarrow \frac{AR}{AD} = \frac{PR}{BD}$ (i) [$\frac{1}{2}$ mark]
Similarly $\Delta ARQ \sim \Delta ADC \Rightarrow \frac{AR}{AD} = \frac{RQ}{DC}$ (ii) [$1$ mark]
Using (i) and (ii) $\frac{PR}{BD} = \frac{RQ}{DC}$ [$\frac{1}{2}$ mark]
AD is the median $\therefore BD = DC \Rightarrow PR = RQ$ [$1$ mark]
i.e. AD bisects PQ [$\frac{1}{2}$ mark]