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Question
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°.
Solution
Given AB || DE, BC || EF
To prove: `∠`ABC + `∠`DEF =180°
Construction: produce BC to intersect 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 angles are supplementary
`∠`DEF + `∠`EML = 180°
From (i) and (ii) we have
∴`∠`DEF + `∠`ABC = 180°
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