Question

$D$ is the midpoint of the side $BC$ of $△ABC$. If $P$ and $Q$ are points on $AB$ and on $AC$ such that $DP$ bisects $\angle BDA$ and $DQ$ bisects $\angle ADC$, then prove that $QP\parallel BC$

Answer 1

In $△ABD,DP$ is the angle bisector of $\angle BDA$.

$\therefore \frac{AP}{PB}=\frac{AD}{BD}$ (angle bisector theorem) .... $\left(1\right)$

In $△ADC,DQ$ is the bisector of $\angle ADC$

$\therefore \frac{AQ}{QC}=\frac{AD}{DC}$ (angle bisector theorem) ...$\left(2\right)$

But, $BD=DC(D$ is the midpoint of $BC)$

Now $\left(2\right)\Rightarrow \frac{AQ}{QC}=\frac{AD}{BD}....\left(3\right)$

From $\left(1\right)$ and $\left(3\right)$ we get,

$\frac{AP}{PB}=\frac{AQ}{QC}$

Thus, $PQ\parallel BC$. (converse of Thales theorem)