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Consider the following right angled triangle with $AB =AC$. $D$ and $E$ are points such that $BD^2 + EC^2 = DE^2$. Prove that $\angle DAE = 45^{\circ}$enter image description here

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?

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  • $\begingroup$ In the first approach, you are trying to prove that $FE=FD \Rightarrow BD=EC$. However, no such condition is imposed - in fact, it could be that $E\equiv C$, and $D$ be the midpoint of $BC$ $\endgroup$ – aaaaa Dec 1 '13 at 5:24
  • $\begingroup$ Okay, so maybe the first approach is too specific. But what about the second one? $\endgroup$ – Gerard Dec 1 '13 at 5:26
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enter image description here

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.

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  • $\begingroup$ Very Nice. By the way with what software did you draw the shape? $\endgroup$ – hhsaffar Dec 1 '13 at 8:20
  • $\begingroup$ @hhsaffar,it is geopad $\endgroup$ – chenbai Dec 1 '13 at 8:23
  • $\begingroup$ Didn't know, Thanks a lot. $\endgroup$ – hhsaffar Dec 1 '13 at 8:24

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