Question

ab and cd are two parallel lines. pq cuts ab and cd at e and f respectively. el is the bisector of​

Answer 1

Direct Answer :

AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of ∠FEB.

Note : I think that the question is not full. So, I corrected it and gave the full question with its solution below -

Full Appropriate Question :

AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of ∠FEB. If ∠LEB =35°, then ∠CFQ will be :

A. 55°

B. 70°

C. 110°

D. 130°

Solution :

Given that,

AB ‖ CD and PQ cuts them

EL is bisector of ∠FEB

∠LEB = ∠FEL = 35°

∠FEB = ∠LEB + ∠FEL

→ 35° + 35°

→ 70°

∠FEB = ∠EFC = 70°ㅤ(Alternate angles)

∠EFC + ∠CFQ = 180°ㅤ(Linear pair)

70° + ∠CFQ = 180°

∠CFQ = 110°

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Answer 2

Question: AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively the bisector of /_FEB. If /_LEB = 35° then /_CFQ will be

Answer:

/_CFQ = 110°

Step-by-step explanation:

Givne that AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively the bisector of /_FEB and /_LEB = 35°.

As EL is the bisector of /_FEB and /_LEB

So,

/_FEB = 2(/_LEB)

/_FEB = 2(35°)

/_FEB = 70°

AB || CD and /_FEB and /_DFE are consecutive interrior angles. So,

/_FEB + /_DFE = 180°

70° + /_DFE = 180°

/_DFE = 180°-70°

/_DFE = 110°

Now,

/_CFQ = /_DFE (vertically opposite angles)

So,

/_CFQ = 110°