The base angles of a triangle are $22.5^\circ$ and $112.5^\circ$. Prove that the base is equal to twice the height of the triangle.
Let the base angles be $B = 22.5^\circ$ and $C = 112.5^\circ$
Draw $AD \perp BC$ extended.
In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$
$\therefore \;$ $A = 180^\circ - \left(B + C\right) = 180^\circ - \left(22.5^\circ + 112.5^\circ\right) = 45^\circ$
In $\triangle ABC$, by sine rule,
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = k$ (say), where $k$ is the constant of proportionality.
$\implies$ $b = \dfrac{a \sin B}{\sin A} = \dfrac{a \sin \left(22.5^\circ\right)}{\sin \left(45^\circ\right)}$ $\;\;\; \cdots \; (1)$
Now, in right $\triangle ACD$,
$\begin{aligned}
AD & = AC \; \sin \left(\angle ACD\right) \\\\
& = b \; \sin \left(180^\circ - \angle ACB\right) \\\\
& = b \; \sin \left(\angle ACB\right) \\\\
& = b \; \sin \left(112.5^\circ\right) \\\\
& = \dfrac{a \times \sin \left(22.5^\circ\right) \times \sin \left(112.5^\circ\right)}{\sin \left(45^\circ\right)} \;\; \left[\text{by equation (1)}\right] \\\\
& = \dfrac{a \times \left(\dfrac{\sqrt{2 - \sqrt{2}}}{2}\right) \times \left(\dfrac{\sqrt{2 + \sqrt{2}}}{2}\right)}{\dfrac{1}{\sqrt{2}}} \\\\
& = \dfrac{\sqrt{2} \times a \times \sqrt{2}}{4} \\\\
& = \dfrac{a}{2}
\end{aligned}$
$\implies$ $a = 2 \; AD$
i.e. $\;$ The base of $\triangle ABC$ is equal to twice its height.
Hence proved.