Question

in given figure prove that AB II EF

Answer 1

$<b >Hey here is your answer$

Given :-

<B = 65°
<BCE = 30°
<ECD = 35°
<CEF = 150°

To Prove :- AB || EF

Proof :-

In figure

<ABC = 65°

<BCD = <BCE + <ECD

= 35° + 30° = 65°

From both of the angles :-

<ABC = <BCD

but these are alternate interior angles on line AB and CD and transversal BC.

so, AB || CD -(1)

NOW ,

<ECD = 30°

<CEF = 150°

By adding both angles

<ECD + <CEF = 30° + 150°
<ECD + <CEF = 180°

but these are cointerior angle on line CD and EF and transversal CE.

so, CD || EF -(2)

Now,

From (1) and (2) :-

AB || EF (hence proved)

hope this helps you

Answer 2

Answer:

<B = 65°

<BCE = 30°

<ECD = 35°

<CEF = 150°

To Prove :- AB || EF

Proof :-

In figure

<ABC = 65°

<BCD = <BCE + <ECD

= 35° + 30° = 65°

From both of the angles :-

<ABC = <BCD

but these are alternate interior angles on line AB and CD and transversal BC.

so, AB || CD -(1)

NOW ,

<ECD = 30°

<CEF = 150°

By adding both angles

<ECD + <CEF = 30° + 150°

<ECD + <CEF = 180°

but these are cointerior angle on line CD and EF and transversal CE.

so, CD || EF -(2)

Now,

From (1) and (2) :-

AB || EF ........provided....