Question

in the given figure AB parallel to CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisector of angle BEF and angle EFD respectively proof that angleEGF is equal to 90 degree ​

Answer 1

let bef = a

as AB //CD

So Dfe = 180-a

EG is bisector of BEF

So FEG = a/2,

And FG is bisector of DFE

So GFE = 180-a)/2 = 90- a/2

In FGE

FGE = 180 - FEG - GFE

= 180 - a/2 - 90 + a/2

= 90

Answer 2

Answer:

In triangle EGF, ∠EGF = 90°.

Step-by-step explanation:

Let ∠BEF = x , ∠DFE = y

EG and FG bisectors.

∠GEF = x/2  ∠GFE =y/2

∠GEF and ∠GFE are Interior angles on the same side of the transversal.

Interior angles on the same side of the transversal are also called as consecutive interior angles or allied angles or co-interior angles.

If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal are supplementary, i.e. they are equal to 180.

here ∠BEF and  ∠DFE are  interior angles.There sum is equal to 180.

therefore ∠BEF + ∠DFE =180°

x + y = 180 .............EQ(1)

considering ΔEGF,

sum of angles in a triangle EGF is equal to 180°.

∠GEF + ∠GFE +∠EGF = 180°

x/2 + y/2 + ∠EGF = 180°

x + y + 2∠EGF = 180*2

from EQ(1)  x + y = 180

180 + 2∠EGF = 180*2

2∠EGF = 180*2 - 180

2∠EGF = 360 - 180

2∠EGF = 180

∠EGF = 180/2

∠EGF = 90°

hence proved.