Question

The midpoint of the side AB of a triangle ABC is D and P is any point on BC. Suppose Q is a point on AC such that ADPQ is a parallelogram. Prove that DQ is parallel to BC.

Answer 1

Given that, in $△ABC,D$ the midpoint of $AB,PandQ$ are points on $BCandAC$ respectively such that, $ADPQ$ is a parallelogram.

To prove: $DQ\parallel BC$

Construction: Join $DQ.$

Proof:

Given that $ADPQ$ is a parallelogram

$AD=PQ$ [Opposite sides of a parallelogram]

$AD=DB$ [$D$ is the mid-point of $AB$]

$\Rightarrow DB=PQ$ $[\because AD=PQ]$

$\Rightarrow DB\parallel PQ$ $[\because PQ\parallel AD]$

$\angle BDP=\angle DPQ$ [Alternate angles]

$DP=DP$ [Common side of $△BDP$ and $△PDQ$]

$\therefore △BDP\cong △QPD$ [By SAS rule]

$\angle BPD=\angle QDP$ [Corresponding angles of congruent triangles]

$\therefore BP\parallel DQ$ [Since alternate angles are equal]

Hence, $DQ\parallel BC$.