How to find an angle between the extension of a diameter and a chord in a semicircle

Question

O is the center of the semicircle, if $\angle EOD = 45^\circ$ and OD = AB, find $\angle BAC$

I don't know what theorem I should use, I've tried to connect many dots to reach the answer but still can't find it. Can anyone show me a hint?

Answer 1

Let $\angle BAC=\alpha$

$\triangle ABO$ is isosceles with $\angle BAO=\angle AOB=\alpha$ and $\angle ABO=180-2\alpha$

$\angle OBE=180-\angle ABO=2\alpha$

$\triangle BOE$ is isosceles with $\angle OBE=\angle BEO=2\alpha$ and $\angle BOE=180-4\alpha$

Now you have $\angle AOB+\angle BOE+\angle EOD=180$

$\alpha+(180-4\alpha)+45=180$

$45-3\alpha=0$

$\alpha=15$