Question

rightsided fig......
class 9th lind and angles
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Answer 1

Solution :

Through C draw CF parallel to both AB and DE.

From this,

• AB || CF

• Transversal BC cuts them at B and C respectively

$\therefore \sf \angle \: ABC \: + \angle \: 1 = {180}^{ \circ}$

[Consecutive interior angles are supplementary)...i)

Similarly,

• DE || CF

• Transversal CD intersects them at C and D respectively

$\therefore \sf \angle \: CDE = \angle \: 2 \:$

[Alternate Angles].....ii)

Adding (i) and (ii), we get :

$\sf \angle \: ABC \: + \angle \: 1 + \angle \: 2 \: = {180}^{ \circ} + \angle \: CDE \:$

$\longrightarrow \: \sf \angle \: ABC \: + \angle \: BCD \: = {180}^{ \circ} + \angle \: CDE \:$

$\because \sf \angle \: 1 + \angle \: 2 = \angle \: BCD \:$

Hence,Proved!