MathLabs

Problem 2

Let II be the incenter of a triangle ABCABC and let Γ\Gamma be its circumcircle. Let the line AIAI intersect Γ\Gamma again at DD. Let EE be a point on the arc BDCBDC (the arc not containing AA) and FF a point on the side BCBC such that ∠BAF=∠CAE<12∠BAC\angle BAF=\angle CAE<\dfrac12\angle BAC. Let GG be the midpoint of the segment IFIF. Prove that the lines DGDG and EIEI intersect on Γ\Gamma.
Step 5 of 6: SAS similarity pins down an equal angle at A' and E
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In plain words

Points AA, II, DD, A′A' all lie on the same ray from AA, so the angle between AFAF and that ray equals the angle between AEAE and the ray; combined with the side ratio from Step 4 this forces triangles FAA′FAA' and IAEIAE to be similar.

∠FAA′=∠IAE  ⟹  △FAA′∼△IAE  ⟹  ∠FA′A=∠IEA\angle FAA'=\angle IAE\implies\triangle FAA'\sim\triangle IAE\implies\angle FA'A=\angle IEA
Detailed analysis

Because II, DD, A′A' all lie on ray AIAI beyond AA, we have ∠FAA′=∠FAI\angle FAA'=\angle FAI and ∠IAE=∠EAI\angle IAE=\angle EAI; Step 1 showed ∠FAI=∠EAI\angle FAI=\angle EAI, so ∠FAA′=∠IAE\angle FAA'=\angle IAE. Together with AFAA′=AIAE\dfrac{AF}{AA'}=\dfrac{AI}{AE} from Step 4, this is exactly the SAS condition for △FAA′∼△IAE\triangle FAA'\sim\triangle IAE under the correspondence F↔I, A↔A, A′↔EF\leftrightarrow I,\ A\leftrightarrow A,\ A'\leftrightarrow E. Similar triangles give ∠AA′F=∠AEI\angle AA'F=\angle AEI, i.e. ∠FA′A=∠IEA\angle FA'A=\angle IEA.