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Problem 2

Consider five points A,B,C,D,EA, B, C, D, E such that ABCDABCD is a parallelogram and BCEDBCED is a cyclic quadrilateral. Let ℓ\ell be a line passing through AA. Suppose that ℓ\ell intersects the interior of the segment DCDC at FF and intersects line BCBC at GG. Suppose also that EF=EG=ECEF = EG = EC. Prove that ℓ\ell is the bisector of angle DABDAB.
Step 1 of 4: Reduce the bisector claim to CF = CG
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In plain words

Alternate interior angles along the parallel sides of ABCDABCD transfer the two halves of ∠DAB\angle DAB to the base angles of △CFG\triangle CFG.

∠GAB=∠GFC,∠FAD=∠FGC  ⟹  (CF=CG  ⟹  ∠GAB=∠FAD)\angle GAB = \angle GFC, \quad \angle FAD = \angle FGC \implies (CF = CG \implies \angle GAB = \angle FAD)
Detailed analysis

Since AB∥DCAB \parallel DC and AD∥BCAD \parallel BC, alternate angles along transversal ℓ\ell give ∠GAB=∠GFC\angle GAB = \angle GFC and ∠FAD=∠FGC\angle FAD = \angle FGC. If CF=CGCF = CG, then △CFG\triangle CFG is isosceles with ∠GFC=∠FGC\angle GFC = \angle FGC, so ∠GAB=∠FAD\angle GAB = \angle FAD and ℓ\ell bisects ∠DAB\angle DAB. Thus it suffices to rule out CF<CGCF < CG and CF>CGCF > CG.