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.
Best Answer
By the law of sines, $$\frac{a}{\sin{45^{\circ}}}=\frac{b}{\sin{22.5^{\circ}}}$$
By the double angle formula, this is equivalent to $$\frac{a}{2\sin{22.5^{\circ}}\cos{22.5^{\circ}}}=\frac{b}{\sin{22.5^{\circ}}}\implies\frac{a}{2\cos{22.5^{\circ}}}=b$$
From the smaller right triangle we see that $$\frac{h}{b}=\cos{22.5^{\circ}}\implies h=b\cos{22.5^{\circ}}$$
Combining the results gives $a=2h$.