Show that the angles satisfy $x+y=z$ How can I show that $x+y=z$ in the figure without using trigonometry? I have tried to solve it with analytic geometry, but it doesn't work out for me. 
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A: Hint: denoting the points like on the picture below, triangles $EGD$ and $DGF$ are similar (why?).

A: Let $X$, $Y$, and $Z$ be the apexes of the angles $x$, $y$, and $z$, respectively.  Also, let $P$ be the common intersection of the red lines.  Show that $ZP^2=ZX\cdot ZY$.  Thus, the circumcircle of the triangle $PXY$ is tangent to $PZ$ at $P$.  This will prove that $\angle YPZ=\angle YXP=x$.
A: Equivalently, you have to prove that
$$
\underbrace{\mathrm{arctan}(1/3)}_{x}+\underbrace{\mathrm{arctan}(1/2)}_{y}=\underbrace{\mathrm{arctan}(1)}_{z}.
$$
This is pretty clear by addition of tangents, indeed
$$
\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}=1 \implies x+y=\frac{\pi}{4}.
$$
A: Imagine that all the the squares are 1 by 1 and so the rectangle has base 3 and a height of $1$. 
There are three right-angled triangles in the diagram. The one with angle $x$ has base 3, and height 1. The triangle with angle $y$ has base 2 and height 1. The triangle with angle $z$ has base 1 and height 1.
Using the standard trig' ratio $\tan \theta = \frac{\mathrm{opp}}{\mathrm{adj}}$, we get $\tan x = \frac{1}{3}$, $\tan y = \frac{1}{2}$ and $\tan z= \frac{1}{1}=1$.
There is a well-know formula for angle addition:
$$\tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$
Applying this formula to the case of $\alpha =x$ and $\beta = y$ gives:
$$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\cdot\frac{1}{2}}=1$$
It follows that $\tan(x+y)=\tan z$. Since $0^{\circ} < x<y<z < 90^{\circ}$ it follows that
$$\tan(x+y) = \tan z \iff x+y = z$$
