Parallelism in A Triangle

Question

Given a triangle $ABC$, $AE$ bisects the angle $BAC$ and $BD$ bisects the angle $ABC$. If $CP$ is perpendicular to $BD$ and $CQ$ is perpendicular to $AE$ then prove that $PQ||AB$.

Answer 1

Let $AE\cap BD=\{O\}$.

Hence, $CPOQ$ is cyclic and $CO$ is a besector of $\angle ACB.$

Thus, $$\measuredangle OQP=\measuredangle OCP=90^{\circ}-\measuredangle POC=90^{\circ}-\left(\measuredangle OBC+\measuredangle OCB\right)=$$ $$=90-\frac{180^{\circ}-\measuredangle BAC}{2}=\frac{\measuredangle BAC}{2}=\measuredangle BAO$$ and we are done!