Question

In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF = 90°.

Answer 1

It is given that, AB || CD and t is a transversal.

∴ ∠BEF + ∠EFD = 180° .....(1) (Sum of the interior angles on the same side of a transversal is supplementary)

EG is the bisector of ∠BEF. (Given)

∴ ∠BEG = ∠GEF = $\frac{1}{2}$∠BEF

⇒ ∠BEF = 2∠GEF .....(2)

Also, FG is the bisector of ∠EFD. (Given)

∴ ∠EFG = ∠GFD = $\frac{1}{2}$∠EFD

⇒ ∠EFD = 2∠EFG .....(3)

From (1), (2) and (3), we have

2∠GEF + 2∠EFG = 180°

⇒ 2(∠GEF + ∠EFG) = 180°

⇒ ∠GEF + ∠EFG = 90° .....(4)

In ∆EFG,

∠GEF + ∠EFG + ∠EGF = 180° (Angle sum property)

⇒ 90° + ∠EGF = 180° [Using (4)]

⇒ ∠EGF = 180° − 90° = 90°