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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 1 of 6: AI bisects angle FAE
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In plain words

Since FF and EE are chosen so that ∠BAF=∠CAE\angle BAF=\angle CAE, the bisector AIAI of ∠BAC\angle BAC automatically bisects the new angle ∠FAE\angle FAE as well.

∠FAI=∠EAI\angle FAI=\angle EAI
Detailed analysis

Since AIAI bisects ∠BAC\angle BAC, we have ∠BAI=∠CAI=12∠BAC\angle BAI=\angle CAI=\tfrac12\angle BAC; subtracting the hypothesis ∠BAF=∠CAE\angle BAF=\angle CAE from these equal halves gives ∠FAI=∠BAI−∠BAF=∠CAI−∠CAE=∠EAI\angle FAI=\angle BAI-\angle BAF=\angle CAI-\angle CAE=\angle EAI, so AIAI also bisects ∠FAE\angle FAE. Moreover DD, the second intersection of AIAI with Γ\Gamma, is the midpoint of arc BDCBDC, the same arc on which EE lies.