Question

In the figure, show that AB || EF

Answer 1

Solution :

In the above figure following things are given :

$\angle{ABC} \: =\: 70^{\circ}$

$\angle{BCD} \: =\: 30^{\circ}\:+\:40^{\circ}$

$\angle{BCD} \: =\: 70^{\circ}$

that means,

$\angle{ABC} \:=\:\angle{BCD}$

Then AB || CD - - - - - (i)

Converse or Postulate

If two angles are interior angle of two parallel lines cut by transversal line have its interior angles whose alternate are also parallel.

$\angle{CEF} \: = \: 140^{\circ}$

$\angle{DCE} \: = \: 40^{\circ}$

$\angle{CEF} \: +\: \angle{DCE} \: = \: 180^{\circ}$

$140^{\circ}\: +\: 40^{\circ}\: =\: 180^{\circ}$

$180^{\circ}\: =\: 180^{\circ}$

Interior angle on same side of transversal

CD || EF - - - - - - (ii)

By equation i & ii

That means

AB || CD || EF

or

$\boxed{\boxed{AB || EF}}$

Hence Proved

Answer 2

Answer:

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Step-by-step explanation:

that it