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In the below fig, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of

∠DEF. Prove that ∠ABC + ∠DEF = 180°.

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#### Solution

Given *AB *|| *DE*, *BC *|| *EF*

To prove: `∠`*ABC *+ `∠`*DEF *=180°

Construction: produce BC to intersect DE at M

Proof: Since *AB *|| *EM *and *B**L *is the transversal

*`∠`ABC *= `∠`*EML [*Corresponding angle*] ……(1)*

Also,

*EF *|| *ML *and *E**M *is the transversal

By the property of co-interior angles are supplementary

*`∠`DEF *+ `∠`*EML *= 180°

From (i) and (ii) we have

*∴`∠`DEF *+ `∠`*ABC *= 180°

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