# Semi similar

DE is diameter of a semicircle with center O. Triangle ADE intersects the semicircle at points B and C. Prove triangle ACB is similar to triangle ADE.

I keep feeling like there should be a simpler and more obvious proof but this is the best I’ve come up with:

$OB = OD \implies \angle OBD = \angle ODB \implies \angle DOB = \pi-2\angle OBD$

$OC = OE \implies \angle OCE = \angle OEC \implies \angle DOC = \angle DOB + \angle BOC = 2\angle OEC$

$OB = OC \implies \angle OBC = \angle OCB \implies \angle BOC = \pi - 2\angle OBC$

So $\pi-2\angle OBD + \angle BOC = 2\angle OEC$

$\implies 2\pi - 2 \angle OBD - 2 \angle OBC = 2 \angle OEC$

$\implies \angle OBD + \angle OBC + \angle OEC = \pi$

But $\angle OBD + \angle OBC + \angle CBA = \pi$

$\implies \angle OEC = \angle CBA$

and $\angle BAC = \angle EAD \implies \triangle ACB \sim \triangle ADE$ QED.