Can I trisect your angle?

It is given that $AB = 4$ and $\angle B = 45^o$ , therefore the altitude of $A$ from $BC$ , $h = 4 \sin{45^o} = 2\sqrt(2)$ .

Since $D$ is the mid-point of $AC$ , the altitude of $D$ is $\frac{1}{2}h = \sqrt{2}$ .

Therefore, the length $BD = \dfrac {\sqrt{2}}{\sin{30^o}} = 2\sqrt{2}$

$\Rightarrow BD^2 = (2\sqrt{2})^2 = \boxed{8}$

Extend segment $AB$ from $B$ . Then locate a point $F$ on that extension such that $BD \parallel FC$ . From this and the Midsegment theorem, we have the following observations:

• Since $BD$ is a median, $AB \cong BF$ and $FC=2BD$ .
• Also, $\angle ABD \cong \angle F$ .

But since $\angle B$ is trisected, we have $\angle ABD=\frac{1}{3} \angle B=15^\circ$ . Thus

• $AB=BF=4$
• $\angle F=\angle ABD=15^\circ$
• $\angle FBC=180^\circ - \angle B=135^\circ$
• $\angle FCB=180^\circ - (\angle F+\angle FBC)=30^\circ$ .

Then sine rule on $\triangle FBC$ gives $\frac{FC}{\sin \angle FBC}=\frac{FB}{\sin \angle FCB} \implies FC=\frac{FB \sin \angle FBC}{\sin \angle FCB}= \frac{4 \sin 135^\circ}{\sin 30^\circ}=4\sqrt 2$ Therefore, $BD=\frac{1}{2}FC=2\sqrt 2$ , and $BD^2=\boxed{8}$ .