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, 9 months ago

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TopNewestOkay,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, 9 months ago

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– Ajala Singh · 3 years, 8 months ago

Thank you!! That really helpsLog in to reply