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1 solution
Let ∠EAD = α. Then ∠AFG = α and also ∠ACB = α. Therefore, ∠CBD = 2α (exterior angle to ΔABC). Also ∠CDB = 2α (since CB = CD). Furthermore, ∠FGC = 2α (exterior angle to ΔAFG). Since GF = EF, ∠FEG = 2α. Now ∠DCE = ∠DEC = β (say). Then ∠DEF = β – 2α. Note that ∠DCB = 180 – (α + β).
Therefore, in ΔDCB, 180 – (α + β) + 2α + 2α = 180 or β = 3α.
Further ∠EFD = ∠EDF = γ (say). Then ∠EDC = γ – 2α. If CD and EF meet at P, then ∠FPD = 180 – 5α (because β = 3α).
Now in ΔPFD, 180 – 5α + γ + 2α = 180 or γ = 3α.
Therefore, in ΔEFD, α + 2γ = 180 or α + 6α = 180 or α = 26 or approximately 25.