My approach is pretty analytical. Let us consider WLOG that the length of the square is 2. Let x be the angle $\angle BAE$. Let us find the equations of the lines AE and BF. For BF, we have the equation $y=mx+b$, and, since $B(2,2)$ and $F(1,0)$ are on the line, we get that $2m+b=2$ and $m+b=0$. Thus, $m=2, b=-2$, so BF has equation $y=2x-2$. For AE, we have the equation $y=mx+b$, and, since $A(0,2)$ and $E(2,1)$ are on the line, we get that $b=2$ and $2m+2=1$, $m=\frac{-1}{2}, b=2$, so y has equation $y=\frac{-1}{2}x+2$. Now we can calculate P's exact location. By writing the intersection condition, we have that $2x-2=\frac{-1}{2}x+2$; by doing routine work, P has coordinates $x_P=\frac{8}{5}$, $y_P=\frac{6}{5}$. As a bonus, we have a proof that PF is perpendicular on AE, which is easily provable through synthetic geometry, too: the slopes of the two lines have product -1, they are perpendicular, so $\angle APF=90$. We shall do a bit of angle chasing now. In the triangles ABE and DCE, $\angle EAB=\angle CDE=x$, $\angle DCE=\angle ABE=90$, so $\angle AEB=\angle DEC=90-x$. Thus, $\angle AED=2x$. Also, $\angle DAP = 90-x$. Back to analytic geometry, we shall look at the lengths of DP and AD. $DP=\sqrt{(\frac{8}{5}-0)^2+(\frac{6}{5}-0)^2}=2=AD$, so the triangle APD is isosceles with base AB. Thus, $\angle DAP=\angle DPA=90-x$, and $\angle PDA=\angle AED=2x$, QED.

Claim.1 : $\angle{BPE} =90$ We have: $$tan(\angle{BAE} ) =\frac{1}{2}=\tan(\angle{EBP}) \Longrightarrow \angle{BAE}=\angle{EBP}$$$$\Longrightarrow \angle{BPE} =90$$. (We can prove it easily with complex numbers.) Claim.2 : $AD$ is a tangent to $(PED)$ Notice that : $$AB^2=AP\cdot AE=AD^2$$$$\Longrightarrow \angle PDA = \angle AED$$$\blacksquare$

Note that $\triangle EAD \cong \triangle FBA$. So, \[\angle PBA = \angle FBA = \angle EAD = \angle AEB.\]So, $\triangle APB \sim \triangle ABE$ which implies \[AP\cdot AE = AB^2 = AD^2\]therefore, $\triangle ADP \sim \triangle AED$. Thus, $\angle ADP = \angle AED$.

Dear Mathlinkers, 1. AE perpendicular to BF and AF perpendicular to DE 2. the tangent to the circle with diameter AF at A is parallel to DE 3. by a converse of the Reim's theorem, the circle trough D, P, E is tangent to AD at D 4. by the Angled tangent theorem, we are done. Sincerely Jean-Louis

$\angle AED=\angle AFB$, obvious, and $AFPD$ cyclic ($AE\bot BF$) solve the problem. Best regards, sunken rock

Let $\angle CDE=x \implies \text { By angle-chasing } \boxed {\angle AED=2x} $ It's easy to observe that $PFDA $ is cyclic Hence, $ \angle PDF=90^{\circ} -2x \implies \boxed{\angle PDA =2x= \angle AED }$

$E, F$ are not necessarily midpoints. It suffices $AE\bot BF.$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20VII.pdf p. 4... Sincerely Jean-Louis

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It's easy to see that $ABE \cong BCF$ and $ADE \cong BAF$. Thus, properties of rotations imply $AE \perp BF$, so $ADFP$ is cyclic by Thales'. Now, we have $$\angle PDA = \angle PFA = \angle BFA = \angle AED$$as desired. $\blacksquare$

Notice $\overline{AE}\perp\overline{BF}$ so $APFD$ is cyclic. Hence, $$\angle PDA=\angle PDA=\angle AED$$by congruency. $\square$