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.

#Geometry #HelpMe! #MathProblem #Math

Note by Ajala Singh
7 years, 6 months ago

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5 votes

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Comments

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

Hence ∠AEF=180∘−∠DEB−∠AEB=75∘\angle AEF = 180^{\circ} - \angle DEB - \angle AEB = 75^{\circ}and by above angle chasing,it is obvious that △AEF∼△CAE\bigtriangleup AEF \sim \bigtriangleup CAE by AAAAAA similarity criteria.I hope this may help.

Kishan k - 7 years, 6 months ago

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Thank you!! That really helps

Ajala Singh - 7 years, 6 months ago
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