We know that AB∥CD and t is a tranversal cutting at points E and F
From the figure we know that ∠BEF and ∠DFE are interior angles
So we get
∠BEF+∠DFE=180o
Dividing teh entire equation by 2 we get
12∠BEF+12∠DFE=90o
According to the figure the above equation can further be written as
∠GEF+∠GFE=90o...(i)
According to the △GEF
We can write
∠GEF+∠GFE+∠EGF=180o
Based on equation (i) we get
90o+∠EGF=180o
∠EGF=90o
Therefore it is proved that ∠EGF=90o