Contruct any angle $\angle O$ and construct its angle bisector. Pick a point on the left leg of the angle and denote it $P$. Drop a perpendicular from $P$ on the angle bisector and denote the point of intersection with the right leg $Q$. Prove that the angle bisector bisects $PQ$.
I have drawn it, it appears to make sense but I can't really see any immediate clever congruences to utilize. Any help?
EDIT: Misread the question, has now been edited.
I don't see how this is true. By your construction $\angle OPQ = 90^{\circ}$, denote $N$ as the bisection point of $PQ$, hence $|PN|=|NQ|$. Suppose for contradiction that the angle bisector bisects $PQ$, then he goes through point $N$. Using Angle bisector theorem we obtain that $\frac{|OP|}{|OQ|}=\frac{|PN|}{|NQ|}=1$ we get that triangle $POQ$ is Right triangle (Because of $\angle OPQ = 90^{\circ}$) and Isosceles triangle (Because $OP=OQ$) then $\angle OQP = 90^\circ$ which only possible if $\angle O = 0^\circ$.
EDIT Let's Denote $N$ as the point of intersection of the perpendicular line with the angle bisector.$ON=ON$,$\angle QNO = \angle PNO = 90^{\circ}$ and $\angle PON =\angle QON$ hence using Angle-Side-Angle theorem we obtain that $\Delta QNO = \Delta PNO$ hence $QN=PN$ as desired.
