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Question

In the given figure, ABCD 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=90o
1715313_5a543b1dea184e0a8aed85f657924d71.png

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Solution

We know that ABCD 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

12BEF+12DFE=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

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