SAT Deja vu

We call $O$ the center of the circle with diameter $AD$ and $G$ the point that satisfies that $\angle DAE = \angle EAG$ , $\angle GDE = \angle EDA$ and $G$ is on the same side of $AD$ as $E$ . Since we know that $AP = PD$ , then $P$ is the midpoint of the semiarc $\overset{\frown}{AD}$ . Similarly, $PO\perp AD$ , and $AO = PO = DO$ , which is the radius of the circle with diameter $AD$ . We know that $B$ is the harmonic conjugate of $D$ with respect to $A$ and $C$ , because $\frac {AB}{BC} = \frac {6}{5} = \frac {66}{55} = \frac {AD}{BD}$ . We also know that $BE\perp ED$ since $BD$ is diameter, then $EB$ is the angle bisector of $\angle AEC$ , so $\angle AEB = \angle BEC = 45°$ , because $AE\perp EC$ since $AC$ is diameter. Using the circle with diameter $AC$ , we notice that $\angle CAE = \angle CFE$ and using the one with diameter $BD$ , we notice that $\angle EDB = \angle EFB$ . Since $\angle CFE + \angle EFB = \angle CFB = \angle BEC = 45°$ , we know that $\angle CAE + \angle EDB = 45°$ . Since we defined $\angle DAE = \angle EAG$ and $\angle GDE = \angle EDA$ , then $\angle CAG + \angle GDB = 2 * (\angle CAE + \angle EDB) = 2 * 45° = 90°$ . Then, $\angle AGD = 90°$ , so $G$ is on the circumference of diameter $AD$ . Since $E$ is the incenter of $\bigtriangleup AGD$ , then $GE$ is the angle bisector of $\angle AGD$ , but since $P$ is the midpoint of the semiarc $\overset{\frown}{AD}$ , then $GP$ is also the angle bisector of $\angle AGD$ , so $G$ , $E$ and $P$ are collinear. We know that $\angle PAO = 45°$ , since $AO = PO$ and $\angle AOP = 90°$ , then $\angle AEP = \angle EAG + \angle AGE = \angle EAG + 45° = \angle DAE + 45° = \angle DAE + \angle PAD = \angle PAE$ . Then, $PA = PE = PD$ , so $PE = PA = \sqrt{AO^2 + PO^2} = \sqrt{33^2 + 33^2} = 46.669$ .

If you let A:(0,0), B:(6,0),C:(11,0) & D: (66,0) then we may obtain E as intersection of (x-11/2)²+y²=(11/2)² and (x-36)²+y²=(30)² which turns out to be (396/61,330/61) while P is (33,-33). From which we can determine PE as 33√2 or ~ 46.669