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# How to do this geometry problem?

I can't make sense of this problem. I can find AEO=30 and OAE=30 so AOE=120. But I can't go further. The diagram is so complicated.

Note by Ajala Singh
3 years, 9 months ago

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Okay,so my solution is like this: Let $$\angle BAC = x$$,therefore $$\angle EAO = 45-x.$$(Diameter subtend $$90^{\circ}$$ at the circumcircle) Also,$$\angle AEB = 30^{\circ}$$,and $$OE=OA$$,so $$45-x=30 \Rightarrow x=15^{\circ}.$$ Notice that $$\angle CBE=\angle CAE \Rightarrow \angle CBE =75^{\circ}.$$ Also observe that quadrilateral $$BCDE$$ is cyclic, therefore $$\angle EDC=105^{\circ} \Rightarrow \angle DEB=75^{\circ}.$$

Hence $$\angle AEF = 180^{\circ} - \angle DEB - \angle AEB = 75^{\circ}$$and by above angle chasing,it is obvious that $$\bigtriangleup AEF \sim \bigtriangleup CAE$$ by $$AAA$$ similarity criteria.I hope this may help. · 3 years, 9 months ago