Question

In the below fig, show that AB || EF.

Answer 1

Produce EF to intersect AC at K.

Now, ∠DCE + ∠CEF = 35° + 145° = 180°

∴ EF || CD    [∵  Sum of Co-interior angles is 180°]    ...(1)

Now, ∠BAC = `∠`ACD = 57°

⇒ BA || CD    [∵ Alternative angles are equal]    ...(2)

From (1) and (2)

AB || EF    ...[Lines parallel to the same line are parallel to each other.]

Hence proved.