How to prove $AP$ is constant in this plane geometry problem?

Question

As is shown in the picture, $ ∠ACB=∠EDB=90^{o},AC=2,BC=4,AE=3,ED=DB,∠PCD=45^{o}$, how to prove that $AP=1$?

By adding additional conditions (like $AE$ is parallel to $CB$) I can verify that $AP=1$. I try to find similar triangles in the graph, but I haven't found.

Answer 1

Let the angle bisector of $ACB$ (which coincidentally gives us $45^\circ$) intersect $AB$ at $F$.
Let $ P^*$ on $AE$ such that $P^* F \parallel EB$.
Let $AC = a$, $CB = b$.

Show the following
(If you get stuck, explain what you've tried and what you're unable to push through)

1. $ \frac{CF}{CB} = \frac{ \sqrt{2} a}{a+b}$
2. $ \frac{P^*F}{DB} = \frac{ \sqrt{2} a}{a+b}$
3. $ \angle P^*FC = \angle DBC$ (angles in red)
4. $ CFP^*$ and $CBD$ are similar triangles.
5. $\angle P^*CF = \angle DCB$ (angles in blue)
6. $\angle P^*CD = 45^\circ$
7. $P = P^*$
8. $\frac{AP}{PE} = \frac{AF}{FB} = \frac{AC}{CB}$

Notes:

• $CP$ (and then $CD$) were the hardest to get a handle on. So we didn't use them directly, but instead got to it from the similar triangles.
• Constructing $F$ was motivated by "knowing" what the answer is, and relating those ratios. This solution was quite natural via a working backwards process.