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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.

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  • $\begingroup$ Sorry but I really can't understand the statement and I can't see the picture. Could you elaborate more? $\endgroup$ – Arnaldo Aug 28 '18 at 20:02
  • $\begingroup$ @Arnaldo thank you for your reply. I tried make it clearer and updated the picture. I hope it helps. $\endgroup$ – B. Schulz Aug 28 '18 at 20:30
  • $\begingroup$ Better now but I still have some questions. So, only the points A, B and T are fixed? $\alpha$ is also constant? BTW, it is not necessary to use Thales to prove that D, A, E are on a circle. They are non-colinear points, so they for a triangle and then there is always a circle through them. $\endgroup$ – Arnaldo Aug 28 '18 at 20:56
  • $\begingroup$ @Arnaldo no $\alpha$ is not constant. $\angle BAC=\alpha$. So $\alpha$ is changing with $C$. $\endgroup$ – B. Schulz Aug 28 '18 at 21:30
  • $\begingroup$ So please, update the statement. That information is not in it. $\endgroup$ – Arnaldo Aug 28 '18 at 21:33
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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 $c$.

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  • $\begingroup$ Thank you for your explanation. Helped me :) $\endgroup$ – B. Schulz Aug 29 '18 at 21:36
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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.

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  • $\begingroup$ Thank you for your different explanation. Helped me also :) $\endgroup$ – B. Schulz Aug 29 '18 at 21:37

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