1.

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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