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.


2 thoughts on “Semi similar

  1. How about this simple explanation?
    From the intersecting chord theorem, AD.AB = AC.AE
    So AD/AE = AC/AB
    so the triangles ADE & ACB share a common angle and the ratio of the lengths of the enclosing sides, and hence are similar.

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