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.

## Comments

Sort by:

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.

Log in to reply

Thank you!! That really helps

Log in to reply