prove that quadrilateral is cyclic 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

 A: 
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.  $\square$
A: Using the same notation, let $H$ be the intersection by extending $AF$ and $DG$. Let the coordinate of $P$ to be $(1+p, 1)$.

It is easy to see that $HFPG$ are concyclic because $\angle F = \angle G = 90^\circ$, and $PH$ is the diameter of the circle, regardless of whether $H$ is between $A$ and $F$. It is left to be proven that $E$ lies on the same circle with $F,P$ and $G$.
The slope of $BP$ is $\frac1{1+p}$, hence the line $AH$ can be written as $$y=1-(1+p)x$$
Similarly, the slope of $CP$ is $-\frac1{1-p}$, hence the line $DH$ can be written as $$y = 1+(1-p)(x-2)$$
Solving for $H$,
$$\begin{align*}
1-(1+p)x &= 1+(1-p)(x-2)\\
-(1+p)x &= (1-p)x-2(1-p)\\
x &= 1-p\\
y &= p^2
\end{align*}$$
If $p=0$, then $H$ is the same as $E$. Otherwise, consider the slopes of $HE$ and $PE$,
$$m_{HE} = \frac{p^2}{(1-p)-1} = -p\\
m_{PE} = \frac{1}{(1+p)-1} = \frac{1}{p}\\
m_{HE} \cdot m_{pe} = -1$$
Hence $HE\perp PE$ and $\angle HEP = 90^\circ$. So $E$ also falls on the circle with diameter $PH$, together with $F$ and $G$.
A: the proof is not pretty, but that is all i can see now.
let us fix some notations:
the length of the rectangle at $2$ and the width $1.$ 
$AP = 1 + x, \angle ABP = \alpha $ and $\angle DCP = \beta$
$F^\prime, G^\prime$ are the feet of the perpendiculars from $F$ and $G$ to $BC.$
you can verify that: $$\tan(\alpha) = 1+x, \tan(\beta) = 1-x, \tan(\alpha+\tan\beta) = {2 \over x^2}.$$
now, $BF = \cos(\alpha), BF^\prime = \cos\alpha \sin\alpha, FF^\prime = \cos^2 \alpha, F^\prime E= 1 - \cos\alpha\sin\alpha$ 
so $$ \tan(\angle BEF) = {\cos^2 \alpha \over 1 - \cos\alpha \sin \alpha} = 
 {1\over 1+\tan^2\alpha - \tan\alpha}= {1 \over 1 + (1+x)^2 -(1+x)} = {1\over1+x+x^2}$$
in the same way $$\tan(\angle CEG) = {1 \over 1 - x + x^2}$$
using the addition formulae for tan gives $$\tan(\angle BEF + \angle CEG) = {2 \over x^2}$$
therefore we now have shown $$\angle BEF + \angle CEG = \angle ABF + \angle DCG = \angle FPG $$ this implies that the opposite angles $FPG$ and $FEG$ sum to $180^\circ$ and that makes the quadrilateral $EGPF$ cyclic.
too much trigonometry and too little geometry.
