Correct Option: A
AB || CD and t transversal intersects them at E and F
∠BEF + ∠EFD = 180° (co-interior angles)
⇒ ∠FEG + ∠EFG = 90°
In Δ GEF
∠EGF + ∠FEG + ∠EFG = 180°
∴ ∠EGF + 90° = 180°
∴ ∠EGF = 90°.
The above result can be restated as :
If two parallel lines are cut by a traversal, then the bisectors of the interior angles on the same side of the traversal intersect each other at right angles.