Question

In figure (10.53), arms BA and BC of ΔABC are respectively parallel to arms ED and EF of Δ DEF. Prove that ΔABC + Δ DEF = 180º.​

Answer 1

Given:-

• AB || DE, BC || EF

To Prove:-

• ∠ABC + ∠DEF = 180º.

construction:-

• produce BC to interest DE at M.

Proof:-

Since AB||EM and BL is the transversal.

∠ABC = ∠EML [corresponding angle] ----------(1)

Also,

EF||ML and EM is the transversal.

By the property of co-interior the angles are supplementary.

∠DEF + ∠EML = 180°----------(2)

From (1) and (2) we have,

∠ABC + ∠DEF = 180°

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