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 $\angle BEF$ and $\angle EFD$ respectivley, prove that $\angle EGF={90}^{\circ }$

Answer 1

ANSWER:
It is given that, AB || CD and t is a transversal.
∴ ∠BEF + ∠EFD = 180°-----(1)
EG is the bisector of ∠BEF.
(Sum of the interior angles on the same side of a transversal is supplementary)
(Given)
∴ ∠BEG = ∠GEF = 12∠BEF
⇒ ∠BEF = 2∠GEF------(2)
Also, FG is the bisector of ∠EFD.
(Given)
∴ ∠EFG = ∠GFD = 12∠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°