Where is my solution wrong for this geometry problem?

Question

2003/4 British Mathematical Olympiad Round 1 Q.2 : "$ABCD$ is a rectangle, $P$ is the midpoint of $AB$, and $Q$ is the point on $PD$ such that $CQ$ is perpendicular to $PD$. Prove that the triangle $BQC$ is isosceles."

My approach: By adding the two equations of Pythagorean theorem of triangle $DQC$ and $PQC$, and taking $DQ = PD - PQ$, we get $CQ^2 = (AB^2 - 2\cdot PQ^2 +2\cdot PD\cdot PQ)/2$.

Now by equation $(-\cos DPA) = \cos QPB = (-AB/2\cdot PD) = [PQ^2 + \{(AB^2)/4\} - BQ^2]/AB\cdot PQ$ (by cosine law in triangle $QPB$)

So we get $BQ^2 = [PQ^2 + \{(AB^2)/4\} + \{AB^2\cdot PQ/2\cdot PD\}]$

But I am not able to show $BQ^2 = CQ^2$.

I would like to have some hints and know about other approaches to solve this problem.

Thank you.

Answer 1

This will be of help to you !! https://artofproblemsolving.com/wiki/index.php?title=Olympiad_Archive

Answer 2

The way without trigonometry and the Pythagoras's theorem:

Let $\measuredangle PDA=\alpha$.

Thus, since $PBCQ$ is cyclic, we obtain: $$\measuredangle BCQ=\measuredangle BCP+\measuredangle PCQ=\alpha+90^{\circ}-2\alpha=90^{\circ}-\alpha=\measuredangle BQC$$ and we are done!