Q181Short answer
In parallelogram ABCD, the angle bisector of ∠A bisects BC. Will angle bisector of B also bisect AD? Give reason.
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Given: ABCD is a gm, bisector of ∠A, bisects BC in F i.e. ∠1 = ∠2, CF = FB BA Const: Draw FE Proof: ABFE is a gm by const. (FE BA) ∠1 = ∠6 (alt. ∠) But ∠1 = ∠2 (given) ∴ ∠2 = ∠6 AB = FB (1) (sides opp to equal ∠s) ∴ ABFE is a rhombus In ∆ABO and ∆BOF AB = BF from (1) BO = BO Common AO = FO Diagonals bisect each other ∆ABO ≅ ∆BOF ∠3 = ∠4 BF = BC (given) BF = AD (BC = AD) AE = AC (BF = AE) ∴ E is mid point of AD