Q11 of 49 Page 151

Let AB, CD be two line segments such that AB || CD and AD ||BC. Let E be the midpoint of BC and let DE extended meet AB in F. Prove that AB = BF.


Given AB || CD and AD || BC hence ABCD is a parallelogram


⇒ AB = CD … opposite sides of parallelogram (i)


Consider ΔCED and ΔBEF


∠DEC = ∠BEF … vertically opposite angles


BE = EC … given


∠ECD = ∠EBF … alternate angles as AF || CD with transversal BC


Therefore, by ASA test for congruency


ΔCED ≅ ΔBEF


⇒ CD = BF … corresponding sides of congruent triangles


Using (i)


⇒ AB = BF


Hence proved


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