Cyclic Quadrilateral Problems Let ABCD be a cyclic quadrilateral. The diagonals AC and BD meet
at P, and DA and CB produced meet at Q. The midpoint of AB is E.
Prove that if PQ is perpendicular to AC, then PE is perpendicular
to BC.
 A: Wow, that was hard.

Let's check that $(\vec{PE}, \vec{CQ})=0$.
$$(2\vec{PE}, \vec{CQ}) = (\vec{PA} + \vec{PB}, \vec{CQ}) = (\vec{PA}, \vec{CQ}) + (\vec{PB}, \vec{CQ}) = (\vec{PA},  \vec{CQ} + \vec{QP}) + (\vec{PB}, \vec{CQ}) = (\vec{PA}, \vec{CP}) + (\vec{PB}, \vec{CQ})$$
In the above equalities the assumption $(\vec{PA}, \vec{QP}) = 0$ was used, if you haven't noticed it.
By power of point theorem, $PA\cdot PC = PB\cdot PD$. Now, let $\angle PAD = \alpha$.
We have $$(\vec{PA}, \vec{CP}) + (\vec{PB}, \vec{CQ}) = PB\cdot PD - PB \cdot CQ \cos \alpha$$ So to show that it is zero it remains to show that $PD\stackrel{?}{=}CQ \cos \alpha$. Let $\angle ADB = \beta$.
By power of point theorem, we have $CQ = \frac{QD\cdot QA}{QB}$, so what we want to prove is $PD\cdot QB \stackrel{?}{=} QD\cdot QA cos \alpha$. From the right triangle $APQ$ we have $QA \cos\alpha = AP$.
$$PD\cdot QB \stackrel{?}{=} QD\cdot AP$$
$$ \frac{QB}{QD} \stackrel{?}{=} \frac{AP}{PD}$$
This is finally true, since from sine theorem for $DBQ$, $\frac{QB}{QD}=\frac{\sin \beta}{\sin \alpha}$ (note that $\angle DBC = \angle DAC = \alpha$) and from sine theorem for $APD$, $\frac{AP}{PD}=\frac{\sin \beta}{\sin \alpha}$. That finishes the proof.
