Question

In the given figure { in attachment } , arms BA and BC of angle ABC are respectively parellal to arms ED and EF of angle DEF. Prove that ,
Angle ABC + Angel DEF = 180°

answer this as per class 9, Lines and Angles

Answer 1

It's given that AB||ED And EF || BC

We'll do a little construction here and extend the lone ED and BC until they intersect at point G

Now,
As EF||BC,

And here BG is is acting as a transversal,
Therefore,
∠ EGB = ∠ DEF (by alternate interior ∠s) -----(i)

Now,
AS AB|| ED
Then,
∠ ABC + ∠ EGB = 180° (interior angles on thr same side of transversal).---(ii)

By substituting for EGB of eq (i) in eqe (ii).

It becomes,
∠ABC+∠DEF= 180°.

Answer 2

Answer:

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Step-by-step explanation:

It's given that AB||ED And EF || BC

We'll do a little construction here and extend the lone ED and BC until they intersect at point G

Now,

As EF||BC,

And here BG is is acting as a transversal,

Therefore,

∠ EGB = ∠ DEF (by alternate interior ∠s) -----(i)

Now,

AS AB|| ED

Then,

∠ ABC + ∠ EGB = 180° (interior angles on thr same side of transversal).---(ii)

By substituting for EGB of eq (i) in eqe (ii).

It becomes,

∠ABC+∠DEF= 180°.