math contest geometry proof problem Could someone help me with this?
Suppose $A,B,C$ are vertices of a triangle and $D$ is a point on the side $BC$. Let $l$ be the line that contains $A$ and bisects $∠CAB$. Suppose there is a point $E$ on $l$ such that upon drawing the line segments $EC$ and $DE$, we have $∠AEC = ∠ABC$ and $∠CDE = 90 ^{\circ}$. Then, show that $|BD|$ = $|CD|$. (Note: $|XY|$ denotes the length of the line-segment $XY$ .)
Drawing a circle around the figure so the base is a chord is what I tried for a while. I couldn't get anywhere else, but you may have more luck with it than I did.
 A: We draw the circumcircle of the triangle $ABC$. The condition states that $∠AEC = ∠ABC$, which means that the point $E$ lies on the circumcircle, beacuse the both angles cut the same chord $AC$. This means that the quadrilaterial $ABCE$ is cyclic.
We draw a normal line to the side $BC$. This line is the line $l$ and cuts the side $BC$ at the point $D$ and the circumcircle of the triangle $ABC$ at point $F$. The quadrilaterial $BCEF$ is cyclic. Here's the proof:
From the quadrialterial $ABCE$ we have $∠BEC + ∠BAC = 180^{\circ}$. Because angles $∠BAC$ and $∠BFC$ lie above the same arch they are equal so this implies $∠BEC + ∠BFC = 180^{\circ}$. And because they are opposite angle it means that $BECF$ is cyclic.
Because the diagonals in the cyclic quadrilaterial $BECF$ are normal to each other it implies that $BECF$ is square (only if ABC is right trinagle) or kite. In any case the diagonal $EF$ cut the other diagonal $BC$ in half.
This leads to $|BD| = |CD|$
Q.E.D.

Here's one eve simplier solution.
Note that for fixed point $B$ and $C$ and fixed circumcircle, the point $E$ will always be on the same position, no matter where $A$ lies on the circumcircle. This is due the fact that the bisector of the $∠BAC$ divides it into two equal angles. Both angles $∠BAE$ and $∠CAE$ are equal and are inscribe angles in the circumcircle, which implies they lie on arches and chords of the same length, i.e $|BE| = |CE|$
Now applying the Pythagorean Theorem on the right triangles $BED$ and $CED$ we have:
$$BD^2 = BE^2 + ED^2 \text{    and    } CD^2 = CE^2 + ED^2 = BE^2 + ED^2$$
This leads to $BD^2 = CD^2$, which implies $|BD| = |CD|$
A: Tip: triangles ABD, CED and ACD are all similar by AA postulate (and ultimately congruent, but that's the proof) Now what do triangles ACD and ABD have in common?
