A geometry problem by A Former Brilliant Member

Let $AB=BC=AD=x$ .

$AC=\sqrt{2x^2}=x\sqrt{2}$ or $x=\dfrac{AC}{\sqrt{2}}$

Extend $CD$ to $E$ so that $AE \bot DE$ . $\angle ADE=x+(45-x)=45$ . It follows that $\angle DAE=45$ .

Since $\triangle AED$ is a $45-45-90$ right triangle, we have

$\dfrac{AE}{x}=\dfrac{1}{\sqrt{2}}$ or $x=\sqrt{2}AE$

Equate $x=x$ :

$\dfrac{AC}{\sqrt{x}}=\sqrt{2}AE$ $\implies$ $AC=2AE$ . Therefore, $\triangle AEC$ is a $30-60-90$ right $\triangle$ .

It follows that $\angle ACE=30$ . Therefore,

$30=45-x$ $\implies$ $x=45-30=$ $\boxed{15^\circ}$

Set $AB=BC=AD=1$ . Then $AC=\sqrt2$ .

From the sum of angles in $\triangle ACD$ the $\angle ADC=180^\circ-x-(45^\circ-x)=135^\circ$

Law of sines for $\triangle ACD$ can be written as $\dfrac{\sin(45^\circ-x)}1=\dfrac{\sin135^\circ}{\sqrt2}$

So that $\sin(45^\circ-x)=\dfrac12$ and $45^\circ-x=30^\circ$

$x=\boxed{15^\circ}$