In the below fig, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF Prove that ∠ABC + ∠DEP = 180°
Question:
In the below fig, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF Prove that ∠ABC + ∠DEP = 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] ... (i)
Also,
EF || ML and EM is the transversal
By the property of co-interior angles are supplementary
∠DEF + ∠EML = 180° .... (ii)
From (i) and (ii) we have
Therefore ∠DEF + ∠ABC = 180°