<s>[INST] Show that base is twice the height if base angles of a triangle are $22.5^\circ$ and $112.5^\circ$ 
The base angles of a triangle are $22.5^\circ$ and $112.5^\circ$. Show that the base is twice the height.

My Attempt

$$
h=c.\sin22.5^\circ=c.\cos 67.5^\circ\\
=b\sin 67.5^\circ=b\cos 22.5^\circ
$$
$$
a=c\cos22.5^\circ- b\sin22.5\circ=\frac{h}{b}-\frac{h}{c}=h\cdot\frac{c-b}{bc}
$$
I have no clue of how to prove this.
 [/INST] The two right angled triangles in your picture are similar and both have smaller angle $22.5$. Let the shortest unmarked side be $x$
Then $$\frac{h}{a+x}=\frac xh=\tan22.5=\sqrt{2}-1$$
Eliminating $x$ gives $$h^2=ah(\sqrt{2}-1)+h^2(\sqrt{2}-1)^2$$
Rearranging gives $$\frac ah=\frac{2\sqrt{2}-2}{\sqrt{2}-1}=2$$
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