TITLE: prove that quadrilateral is cyclic
QUESTION [7 upvotes]: let $ABCD$ be a rectangle with $BC=2AB$.Let E be the midpoint of side BC and P an arbitrary inner point of AD. Let F and G be the feet of perpendiculars drawn from A to BP and from D to CP. Prove that E,F,P,G are concyclic.
I have tried this problem a lot to solve but couldn't. I dont want full solution but any hint will be highly appreciated. Thanks

REPLY [4 votes]: Note that 
$$\frac{|\overline{BF}|}{|\overline{BA}|} = \frac{|\overline{BA}|}{|\overline{BP}|} \quad\to\quad |\overline{BF}||\overline{BP}| = |\overline{BA}|^2 = |\overline{BE}|^2\quad(\star)$$
Let $\gamma$ be the circle through $P$, tangent to $\overline{BC}$ at $E$. The product $|\overline{BE}|^2$ gives the "power of a point" for $B$ with respect to $\gamma$. Since $(\star)$ shows that $|\overline{BF}||\overline{BP}|$ gives this power as well, and since $P$ is on $\gamma$, we may conclude that $F$ is on $\gamma$, too. Similar reasoning, based on the power of $C$ with respect to $\gamma$, shows that $G$ is also on the circle.  $\square$