A friend of mine presented the following problem to me: Say you have a square $ABCD$ and a kite $BCEF$ (a kite is a quadrilateral object such that $\overline{BC}$ and $\overline{BF}$ have the same size, $\overline{CE}$ and $\overline{EF}$ have the same size and $\overline{BE}$ and $\overline{CF}$ are perpendicular), much like in the picture bellow (also, sorry for the picture, I cannot figure out how to make such image in the latex editor of the site).
If you prolong the segments $\overline{BE}$ and $\overline{AD}$, they should meet in some point $H$. Likewise, the prolongation of $\overline{BF}$ meet with $\overline{AD}$ in $G$.
The problem: Show that $a + b = c$.
I cannot enphasize how mad I am at this problem. I know that the triangles $DEH$, $BCE$ and $BEF$ are equivalent, and I've spent some time walking in circles, nothing really that useful.
If someone could point out what I am not seeing, I would be deeply grateful.
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$\begingroup$Green coloured angles are equal and let them be $x$. $\triangle BGH$ is isosceles. Therefore $BG=GH=c$.
Now rotate $\triangle BCE$ about point $B$, $90$ degrees clockwise to make $\triangle BAE'$.
Then $E'G=b+a$.
$\angle E'BH=90^\circ$, so $\angle E'BG=90^\circ-x$.
Considering the triangle $BCE$, $\angle BEC=90^\circ-x$ and so is $\angle BE'G$.
Hence, $\angle BE'G=\angle E'BG$, then $\triangle BE'G$ is isosceles and $E'G=BG$ $\implies$ $$a+b=c$$
$\endgroup$ 2 $\begingroup$Let $GD=x$
$\triangle CBE \sim \triangle DHE$
$$\frac{a+x}{c-x}=\frac{b}{a+x-b}$$
$$(a+x)^2-b(a+x)=b(c-x)$$
$$c^2-a^2-b(a+x)=b(c-x)$$
$$c^2-a^2=b(c-a)$$
Since $a\neq c$
$$c=a+b$$
Since $\triangle ABG$ is a right triangle.
$$a^2+(a+x)^2=c^2$$
$\endgroup$ $\begingroup$Let $\alpha$ be the common measure of the angles $\widehat{CBE}$ and $\widehat{EBF}$. Let $t$ be $t=\tan\alpha$. We may and do assume that the four sides of the square have each the length one. Let us show the equality: $\color{blue}{2a+b=a+c}$. We have:$$ \begin{aligned} b &= \tan \alpha = t\ ,\\ a &= \tan(90^\circ-2\alpha)=\cot (2\alpha)=\frac{1-t^2}{2t}\ ,\\ c+a &= \tan(90^\circ-\alpha)=\cot \alpha=\frac1t\ ,\qquad\text{so}\\ \color{blue}{2a +b}&= 2\cdot\frac{1-t^2}{2t} +t =\frac 1t =\color{blue}{c+a} \ . \end{aligned} $$
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