Proving $AE+AP=PD$ In a Certain Right Triangle 
$\angle B$ in $ \triangle ABC$ is right. 
The incircle of $ \triangle ABC$ is tangent to the side $AB,BC,CA$ in $E,D,F$. 
The line $AD$ meets the incircle of $ \triangle ABC$ in $P(\neq D)$. 
If $\angle BPC$ is right, prove that $AE+AP=PD$.
I have managed a algebraic solution to the problem, seen here, which I obtained by drawing the excircle of $ \triangle BPC$, which met the line $AP$ in $X(\neq P)$
However, I am currently unable to manage a geometric solution. Any introduction to a geometric approach would be appreciated. 
 A: Very nice problem! Although $\angle B=90^{\circ}$ is unnecessary. 
Let $PB,PC$ meet the incircle again at $X,Y$. Let $K=EF\cap BC$. Since $K$ lies on the polar of $A$ wrt $(I)$, it follows that $AD$ is the polar of $K$, so $AP$ is tangent to $(I)$. Now considering the complete quadrilateral $\{AB,BC,CA,EF\}$, $$(P,D;X,Y)\stackrel{P}{=}(K,D;B,C)=-1,$$ so it follows that $P,D$ are symmetric in $XY$.
Hence $\triangle PYD$ is $Y$-isosceles. Let $M$ be the midpoint of $\overline{DE}$. Since $PY$ is the $P$-symmedian of $\triangle PED$, it follows that $\triangle PYD\sim\triangle PEM$, so $PE=\frac{1}{2}ED$. Also, since $\triangle AED\sim\triangle APE$, we get $AE=2AP$ and $AD=2AE=4AP$, so $AE+AP=3AP=PD$ as required.
A: Chad Shin:  after your kind comment about the triangle defined by $A=(0,4+\sqrt 7), B=(0,0),C=(18+6\sqrt 7,0)$  I realize (I can be wrong, it is a personal feeling) that your post is less relevant than I had thought. It was not a property but just an arithmetic result of a particular case. 
Let me, please, explain my reasoning now about your problem.
In the picture below, Figures $1$, $2$ and $3$ have the same fixed red circle $\Gamma$ which gives fixed points $E$ and $D$. 

In Figure $1$, the arbitrary point $C$ determines by tangency to $\Gamma $ the triangle $\triangle ABC $ of which $\Gamma $ is the incircle. This determines the point $P$ as equal to $\Gamma\cap\overline{AD}$ .
In addition, the green semicircle centered at the midpoint of the segment $\overline {BC}=a$  and of radius $\frac a2$ contains all interior points $Q$ of $\Gamma$ such that $\angle BQC=90^{\circ} $, in particular   the point $P’$, in which this semicircle intersects the line AD.
The fact to be underlined is that for all (red) circle fixed, arbitrary points $C$ determine both points $P$ and $P '$. And these points generally do not coincide.
Look now at the Figure $2$ where the point $C$ has been chosen to the left of the correlative $C$ in figure $1$ so the blue semicircle is smaller than the green one: in this case the segments $\overline {PP’}$, $\overline {AE}$ and $\overline {AP}$  are larger than their correlative in Figure $1$.
On the contrary, in Figure $3$ (with purple semicircle defining $P'$) where the point $C$ has been chosen to the right of the correlative $C$ in Figure $1$, the segments $\overline {PP’}$, $\overline {AE}$ and $\overline {AP}$ are smaller than their correlative in Figure $1$.
It is not very formal (I don’t want to try and it seems easy enough) but it is true by continuity that for the same arbitrary red fixed circle there is a position of the point $C$ such that $\overline {AP}+\overline {AE}$ equals all value between its minimum and its maximum, in particular could be equal to $\overline {PD}$ but the same conclusion is valid for $\overline {AD}$ instead of  $\overline {PD}$ and for any segment whose length is between the maximum and the minimum allowed to the red circle fixed.
Your triangle defined by $A=(0,4+\sqrt 7)$,  $B=(0,0)$,  $C=(18+6\sqrt 7, 0)$ of which you say “This appears to be the only triangle in which all conditions are met”, I believe without verification you are right on satisfied conditions. However your conclusion of unicity is valid but referred to the semicircle of radius $9+3\sqrt 7$ and (red) incircle of radius $r=\frac {79+42\sqrt 7}{599+224\sqrt 7}$ (except miscalculations).  And there are infinitely many possible red circles to consider.
