Question

In Fig 8.108, show that AB|| EF.

Answer 1

Given :  From figure : ∠ECD  35°, ∠ACE = 22°  , ∠CEF =  145° .

To Prove  :  AB || EF

Proof:  

∠ACD = ∠ACE + ∠ECD

∠ACD = 22° + 35°

∠ACD = 57°

∴ ∠BAC = ∠ACD

Thus lines AB and CD are intersected by the line BC such that ∠ABC = ∠BCD i.e. , the alternate angles are equal.

∴ AB || CD ………..(1)

Now,

∠DCE + ∠CEF =  35° + 145°

∠DCE + ∠CEF = 180°

This shows that the sum of the interior angles on the same side of a transversal CE is 180° i.e., they are supplementary.

∴ CD || EF …………(2)

From eq 1 and 2 , we obtain  

AB || CD and CD || EF  

∴ AB || EF

Hence, proved

HOPE THIS ANSWER WILL HELP YOU…..

Some questions of this chapter :

In Fig 8.112, if l||m, n||p and ∠1 =85°, find ∠2.

https://brainly.in/question/15905987

In Fig 8.114, AB||CD and P is any point shown in the figure. Prove that:

∠ABP + ∠BPD + ∠CDP = 360°

https://brainly.in/question/17047019

Answer 2

Step-by-step explanation:

______❤HEY! MATE ❤_____

HERE IS YOUR ANSWER! !!!!

FIRST OF ALL,

<BAC =57°

<ACE=22°

<ECD= 35°

NOW,

<ACE + <ECD = 22° +35° = 57°

HENCE,

<BAC = < ACD ( ALTERNATE ANGELES )

SO,

AB||CD---------(1)

NOW,

<FEC = 145°

<FEC + < ECD = 145°+35° =180°

HENCE THEY ARE CO- INTERIOR ANGLES.

NOW, CD||EF ---------(2)

FROM EQ 1 AND 2 WE CAN SAY THAT.

AB||EF

HENCE PROVED.