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Question

In the given figure, $$AB\parallel CD$$ and a transversal $$t$$ cuts them at $$E$$ and $$F$$ respectively. If $$EG$$ and $$FG$$ are the bisectors of $$\angle BEF$$ and $$\angle EFD$$ respectively, prove that $$\angle EGF={90}^{o}$$
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Solution

We know that $$AB\parallel CD$$ and $$t$$ is a tranversal cutting at points $$E$$ and $$F$$

From the figure we know that $$\angle BEF$$ and $$\angle DFE$$ are interior angles

So we get

$$\angle BEF+\angle DFE={ 180 }^{ o }$$

Dividing teh entire equation by 2 we get

$$\dfrac{1}{2}\angle BEF+\dfrac{1}{2}\angle DFE={90}^{o}$$

According to the figure the above equation can further be written as
$$\angle GEF+\angle GFE={90}^{o}...(i)$$

According to the $$\triangle GEF$$

We can write

$$\angle GEF+\angle GFE+\angle EGF={ 180 }^{ o }$$

Based on equation (i) we get

$${90}^{o}+\angle EGF={ 180 }^{ o }$$

$$\angle EGF={90}^{o}$$

Therefore it is proved that $$\angle EGF={90}^{o}$$

Mathematics
RS Agarwal
Standard IX

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