Because $\angle CAE = \angle ADE + \angle AED = \angle AFC + \angle ACF = \angle FAD$ and ${l_1} \bot CD$, ${l_1}$ is the bisector of $\angle EAF$, denote ${l_1} \cap \left( {\Delta AEF} \right) = X$, then $XE = XF$, so $X = {l_1} \cap {l_2} = P$. Because $\angle EBF = \angle AEB + \angle AFB + \angle EAF = \angle BCD + \angle BDC + \angle EAF$, and $\angle EBF = \angle 180^\circ - \angle BCD - \angle BDC$, plus them we have$2\angle EBF = 180^\circ + \angle EAF = 360^\circ - \angle EPF$, so $P$ is the circumcenter of $\Delta EBF$. Hence $D{P^2} - P{E^2} = DB \cdot DE = DA \cdot DC = 2A{C^2}$, because $D{P^2} = A{D^2} + A{P^2} = A{C^2} + A{P^2}$, we have $A{P^2} = P{E^2} + A{C^2}$.

very nice solution! Thank you. But, i have a other idea. Let $k_1$ - circle with center $A$ and radius $AC$, $k_2$ - circle with center $P$ radius $PE$. Idea: $k_1$ and $k_2$ - orthoganal(intersecting at right angle). This idea is beautiful, but i stopped, here. Please, help me!

What is meant by l_1 and l_2 have a unique common point? Don't all pairs of lines have a unique common point?

Generalization: Two circles $ \mathcal{K}_1 $ and $ \mathcal{K}_2 $ of different radii intersect at two points $ A $ and $ B $, let $ C \in \mathcal{K}_1 $ and $ D \in \mathcal{K}_2 $ satisfy $ A \in CD $ . The extension of $ DB $ meet $ \mathcal{K}_1 $ at another point $ E $, the extension of $ CB $ meet $ \mathcal{K}_2 $ at another point $ F $ . Let $ P $ be the intersection of a line passing through $ A $ which is perpendicular to $ CD $ and the perpendicular bisector of $ EF $ . Prove that $ AP^2=EP^2+AC \cdot AD $ Proof: Let $ X=AP \cap EF $ . Since $ \triangle ACE \sim \triangle AFD $ , so we get $ AC \cdot AD=AE \cdot AF $ . ... $ (1) $ Since $ \angle EAP=90^{\circ}-\angle CAE=90^{\circ}-\angle FAD=\angle PAF $ , so $ P $ is the intersection of the bisector of $ \angle EAF $ and the perpendicular bisector of $ EF $ , hence we get $ P $ is the midpoint of arc $ EF $ in $ (AEF) $ and $ EP^2=XP \cdot AP $ . ... $ (2) $ Since $ \triangle AEX \sim \triangle APF $ , so we get $ AX \cdot AP=AE \cdot AF $ . ... $ (3) $ From $ (1), (2), (3) $ we get $ EP^2+AC \cdot AD=XP \cdot AP+AX \cdot AP=AP^2 $ . Q.E.D

TelvCohl wrote: Let $ P $ be the intersection of the perpendicular bisector of $ CD $ and the perpendicular bisector of $ EF $ . I think you mean the line perpendicular to $CD$ through $A$