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In the given figure, AB CD and a transversal t cuts them at E

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In the given figure, $A B ∥ C D$ and a transversal $t$ cuts them at $E$ and $F$ respectively. If $EP$ and $FQ$ are the bisectors of $∠ A EF$ and $∠ EF D$ respectively, prove that $EP ∥ FQ$

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Emma Discussed

In the given figure,

A B ∥ C D

and a transversal

t

cuts them at

E

and

F

respectively. If

EP

and

FQ

are the bisectors of

∠ A EF

and

∠ EF D

respectively, prove that

EP ∥ FQ

10 mins ago

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10 mins ago

Text solutionVerified

We know that

A B ∥ C D

and

t

is a transversal

from the figure we know that

∠ A EF

and

∠ EF D

are alternate angles

So we get

∠ A EF = ∠ EF D

Dividing both side by 2 we get

\frac{1}{2} ∠ A EF = \frac{1}{2} ∠ EF D

So we get

∠ PEF = ∠ EFQ

The alternate interior angles are formed only when the transversal

EF

cuts both

FQ

and

EP

Therefore it is proved that

EP ∥ FQ

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Question Text	In the given figure, $A B ∥ C D$ and a transversal $t$ cuts them at $E$ and $F$ respectively. If $EP$ and $FQ$ are the bisectors of $∠ A EF$ and $∠ EF D$ respectively, prove that $EP ∥ FQ$
Answer Type	Text solution:1
Upvotes	150

In the given figure, AB∥CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP∥FQ

Text solutionVerified

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In the given figure, $A B ∥ C D$ and a transversal $t$ cuts them at $E$ and $F$ respectively. If $EP$ and $FQ$ are the bisectors of $∠ A EF$ and $∠ EF D$ respectively, prove that $EP ∥ FQ$