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 parallel to BC, by similarity and proportionality theorem in triangles APR and ABD, and ARQ and ADC, we get $\frac{A R}{A D} = \frac{PR}{B D}$ and $\frac{A R}{A D} = \frac{RQ}{D C}$ . Since AD is median, BD=DC, hence PR=RQ, meaning AD bisects PQ.

Step-by-step working

1
Given $PQ ∥ BC \Rightarrow PR ∥ B D$ . Therefore in $△ A PR$ and $△ A B D$ , $∠ P A R = ∠ B A D$ and $∠ A PR = ∠ A B D$ . $△ A RP \sim △ A D B$ (By AA similarity criterion). $\Rightarrow \frac{A R}{A D} = \frac{PR}{B D}$
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2
Similarly $△ A RQ \sim △ A D C$ . $\Rightarrow \frac{A R}{A D} = \frac{RQ}{D C}$
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3
Using above: $\frac{PR}{B D} = \frac{RQ}{D C}$
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4
AD is the median $∴ B D = D C \Rightarrow PR = RQ$
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5
i.e. AD bisects PQ
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Concepts used

medianparallel linessimilar trianglesproportionality

Approach

Step-by-step working

Concepts used

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