TITLE: In this isosceles right angled triangle, prove that $\angle DAE = 45^{\circ}$
QUESTION [5 upvotes]: Consider the following right angled triangle with $AB =AC$. $D$ and $E$ are points such that $BD^2 + EC^2 = DE^2$. Prove that $\angle DAE = 45^{\circ}$
The obvious thing was to construct a right angled triangle with $BD, EC, DE$ as its sides. So, we draw a circle around $BD$ with $D$ as centre and a circle around $EC$ with $E$ as centre. Let the intersection of the circle in the interior of the triangle be $F$. Join $FD$ and $EC$. Clearly, $DF = BD$ and $FE = EC$. Also, $\angle DFE = \frac{
\pi}{2}$
If we can prove that $F$ is the circumcenter of $\Delta DAE$, we are done, since the angle at the centre is double the angle anywhere else on the circle and hence $\angle DAE$ has the desired measure. We can prove this by proving that $AF = FE = FD$.
Another way to approach the problem is: 
Let $\angle FED = \theta$. 
$$\implies \angle FEC = 180 - \theta$$ 
$$\angle EFC = \angle ECF = \frac{\theta}{2}$$
$$\angle EDF = 90 - \theta$$
$$\angle BDF = 90 + \theta$$
$$\angle DBF = \angle DFB = 45 - \frac{\theta}{2}$$
If we can prove that $BDFA$ and $FECA$ are cyclic, $\angle FCE = \angle FAE = \frac{\theta}{2}$. $\angle DBF = \angle DAF = 45 - \frac{\theta}{2}$. Summing up the values of $\angle DAF$ and $\angle FAE$ gives us the desired value. 
I have not been able to prove the required things for each of these approaches. How do I prove them?

REPLY [3 votes]: The approach is to prove K is the circumcenter of △ADE not J or your F.
let $BD=x,CE=y,DE=\sqrt{x^2+y^2}=2r,BC=d=x+y+2r$
$N$ is mid point of $DE,KN=NE=ND=r,KD=KE=\sqrt{2}r$
$AK^2=KH^2+AH^2$
$KH=MN=\dfrac{d}{2}−x−r=\dfrac{y−x}{2},AH=\dfrac{d}{2}−r=\dfrac{x+y}{2},AK^2=\dfrac{x^2+y^2}{2}=2r^2 \implies AK=\sqrt{2}r=KE=KD$
so $K$ is the circumcenter of $△ADE$
then $∠DAE=45^{\circ}$ is trivial.