Triangle Point stays the same The Point $C$ of the triangle $ABC$ lies on the perpendicular to $AB$ trough Point $T$. $\alpha$ is the angle $\angle BAC$. Point $D$ lies on a line with the $\angle CBD=\alpha$. The Point $E$ lies on the perpendicular to $AC$ trough $D$. Proof that $E$ is on $AB$ and stays the same, while moving $C$ on the perpendicular.
I prooved that $D$, $A$ and $E$ are on a Circle by using Thales's theorem (like in this picture). But now I stuck.
 A: Set $AT = a$, $BT = b$, and $CT = c$. Specifying these lengths fully specifies the picture; all we have to do now is prove that $AE$ depends on $a$ and $b$, but not $c$.
We can find $AC$ and $BC$ using the Pythagorean theorem. Because $\triangle ABC$ and $\triangle BDC$ are similar (two of their angles are equal), we can use equal ratios to solve for $CD$ in terms of $AC$ and $BC$; since $AC = AD + CD$, we can solve for $AD$ as well.
The second pair of similar triangles is $\triangle ACT$ and $\triangle AED$: they share $\angle A$ and are both right triangles. So $\frac{AE}{AD} = \frac{AC}{AT}$, and we can solve for $AE$.
If you do the arithmetic correctly, you should get $AE = \frac{a^2-b^2}{a}$, which does not depend on $c$.
A: The quadrilateral $ETCD$ is inscribed since it has right angles at $D$ and $T$. So $AE\times AT=AD \times AC=AC^2-CD\times CA$. The circle through $A,D,B$ is tangential to $CB$ since $\angle DAE= \angle CBD$, so $CD \times CA=CB^2$. Combining these identities give $AE \times AT=AC^2-CB^2=AT^2-TB^2$ by the pythagorean theorem. It follows that $AE \times ET$ does not depend on $C$ and so $E$ is a fixed point. 
