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Question
In the given figure, `square`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.
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Solution
`square`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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