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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, 11 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.

Kishan K - 3 years, 11 months ago

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Thank you!! That really helps

Ajala Singh - 3 years, 10 months ago

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