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 $$EP\parallel FQ$$

Solution

## We know that $$AB\parallel CD$$ and $$t$$ is a transversalfrom the figure we know that $$\angle AEF$$ and $$\angle EFD$$ are alternate anglesSo 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 $$EP\parallel FQ$$MathematicsRS AgarwalStandard IX

Suggest Corrections

0

Similar questions
View More

Same exercise questions
View More