

InterviewSolution
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In the adjoining figure, □ABCD is a parallelogram. Point E is on the ray AB such that BE = AB, then prove that line ED bisects seg BC at point F. |
Answer» Given: □ABCD is a parallelogram. BE = AB To prove: Line ED bisects seg BC at point F i.e. FC = FB Proof: □ABCD is a parallelogram. [Given] ∴ seg AB ≅ seg DC …….(i) [Opposite angles of a parallelogram] seg AB ≅ seg BE ……..(ii) [Given] seg DC ≅ seg BE ……..(iii) [From (i) and (ii)] side DC || side AB [Opposite sides of a parallelogram] i.e. side DC || seg AE and seg DE is their transversal. [A-B-E] ∴ ∠CDE ≅ ∠AED ∴ ∠CDF ≅ ∠BEF …..(iv) [D-F-E, A-B-E] In ∆DFC and ∆EFB, seg DC = seg EB [From (iii)] ∠CDF ≅ ∠BEF [From (iv)] ∠DFC ≅ ∠EFB [Vertically opposite angles] ∴ ∆DFC ≅ ∆EFB [SAA test] ∴ FC ≅ FB [c.s.c.t] ∴ Line ED bisects seg BC at point F. |
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