CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In the given figure, $$AB\parallel CD$$ and a transversal $$t$$ cuts them at $$E$$ and $$F$$ respectively. If $$EP$$ and $$FQ$$ are the bisectors of $$\angle AEF$$ and $$\angle EFD$$ respectively, prove that $$EP\parallel FQ$$
1715317_1e11e70a16644d89921ec1271d2416e3.PNG


Solution

We know that $$AB\parallel CD$$ and $$t$$ is a transversal
from the figure we know that $$\angle AEF$$ and $$\angle EFD$$ are alternate angles

So we get

$$\angle AEF=\angle EFD$$

Dividing both side by 2 we get

$$\dfrac{1}{2}\angle AEF=\dfrac{1}{2}\angle EFD$$

So we get

$$\angle PEF=\angle EFQ$$

The alternate interior angles are formed only when the transversal $$EF$$ cuts both $$FQ$$ and $$EP$$

Therefore it is proved that $$EP\parallel FQ$$

Mathematics
RS Agarwal
Standard IX

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More



footer-image