As many problems with special configuration, the analysis of construction is of paramount importance for this problem as well. Suppose we have D given on (AB); draw DD' || BC (D' on AC) and construct the symmetrical of AC about the perpendicular bisector of DD', it will intersect BC at E. P will be on this perpendicular bisector. From the similar triangles DEP and PCA we get DP/PA = DE/PC = PE/AC or (DP/PA)^2 =(DE/PC).(PE/AC), but PE = PC and it remains (DP/PA)^2 = DE/AC ( 1 ). Applying the sine theorem to triangles BDE and ABC with m(<BED) = 180 - m(<ACB) ( from construction ) we get BD/AB = DE/AC ( 2 ), but the tangent at P to the circle (ADP) will intersect AD at a point T with property TD/TA = (DP/AP)^2 ( 3 ). From (1), (2) and (3) we get T=B. Best regards, sunken rock

To avoid using the sine theorem to get BD/AB = DE/AC: Name R the intersection of AC and DE, apply Menelaos theorem to triangle ADR with the transversal BEC, see that ER = CR. Best regards, sunken rock

(Sorry if the post is old, but I think I've found another solution with no computation?) Reflect A over the perpendicular bisector of EC, call this point F; note that EP must be tangent to the circumcircle of DPF by angle condition. Invert the diagram about P: We get that still PA' = PF' and PC' = PE'. The point D' is such that PF'D'E' is an isoceles trapezoid with PF' = E'D'. It suffices to show that if B' is the point such that PB'D'A' is an isoceles trapezoid (B'D' = PA'), then C'EB'P lie on a circle, so consider this point B'. Let O be the intersection of the perpendicular bisectors of PB' and PE'. We have OA' = OD' = OF' so O is also on the perpendicular bisector of E'C' and so we have found O is the circumcenter of E'B'PC' and we are done.

Hello. I think that this is a nice problem. I had solved it in the way of sunken rock. First off, one way to "complete the diagram" in some sense is to add the intersection of $ DE$ and $ AC$ (call it it $ X$). Now you have two isosceles triangles. If we decide to start with drawing $ XCE$, $ PCE$, then it is trivial to find $ A,D$ and then construct $ B$ as the intersection of $ CE$ and $ DA$. If we look at allen's solution, we note that his point "F" lies on $ XE$ as I've defined it. Furthermore, another way to motivate the use of inversion is that if we invert about $ P$, the similar triangles that are so difficult become congruent! This important observation takes the two odd assumptions of the triangle similarity and the equal lengths and makes it more reasonable to manage. I also am curious, how did you come up for the construction of point $ F$? It is very nice in the fact that you have two isosceles trapezoids that share a diagonal the diagonal $ PD$. Also the construction of $ O$ is nice...

My motivation for F was basically what you said - since the angles DEC and ACE are equal, it's natural to extend DE and AC to meet at X, then it just makes a lot of sense to reflect A across the perpendicular bisector of EC, since you really don't have to worry about the original A that much anymore. Then well, inversion looks too nice with the tangent circles to lines. The construction of O was mostly from all that reflective symmetry of isosceles triangles and trapezoids.

My solution : Let $ A' $ be the reflection of $ A $ in the perpendicular bisector of $ CE $ . Let $ R $ be the radius of $ \odot (ADA') $ and $ O $ be the circumcenter of $ \triangle ADA' $ . From $ \angle CEA'=\angle ACE=\angle CED $ $ \Longrightarrow $ $ A', D, E $ are collinear . From $ \triangle DEP \sim \triangle PCA $ $ \Longrightarrow $ $ EP^2=EP \cdot CP=ED \cdot CA $ . i.e. $ EP^2=EO^2-R^2 $ From $ OP \perp BE $ $\Longrightarrow $ $ BO^2-BP^2=EO^2-EP^2=R^2 $ , so $ BD \cdot BA=BO^2-R^2=BP^2 $ . i.e. $ BP $ is tangent to $ \odot (PAD) $ at $ P $ Q.E.D

Let the tangent to $\odot (PAD)$ at $P$ meet $AD$ at $B'.$ Then from $\triangle B'PD \sim \triangle B'AP$, we obtain \[\frac{PD^2}{PA^2} = \frac{B'P^2}{B'A^2} = \frac{B'D \cdot B'A}{B'A^2} = \frac{B'D}{B'A}.\] Meanwhile, note that $\angle PED = \angle PCA$ and $\angle PEC = \angle PCE.$ Consequently, $\angle BED = 180^{\circ} - \angle DEC = 180^{\circ} - \angle ACB.$ Now let $F \ne C$ lie on $BC$ with $AF = AC.$ Then it follows that $DE \parallel AF$, so from $\triangle BDE \sim \triangle BAF$, we get \[\frac{BD}{BA} = \frac{DE}{AF} = \frac{DE}{AC}.\] From $\triangle DEP \sim \triangle PCA$ we have \[\frac{DE}{AC} = \frac{DE / PE}{AC / PC} = \frac{DE^2}{PC^2} = \frac{PD^2}{PA^2}.\] Therefore, $\tfrac{BD}{BA} = \tfrac{B'D}{B'A}$, and the desired result follows because both $B$ and $B'$ lie in the exterior of $\overline{AD}.$