In the figure given, parallel lines, AB and CD, are intercepted by a transversal t at E and F, respectively. If EG is the angle bisector of $\angle AEF$, find $\angle EGF$.
∠AEF = Corresponding ∠CFt (external) = 70°.
∠GEF = ∠AEF ÷ 2 = 70° ÷ 2 = 35°.
∠EFG = 180° − (70° + 25°) = 85°.
∠EGF = 180° − (35° + 85°) = 180° − 120° = 60°.
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