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Question
In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.
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Solution
In Δ BEF and ΔCED
`∠`BEF = `∠`CED [Verified opposite angle]
BE = CE [ ∵ E is the mid-point of BC]
`∠`EBF = `∠`ECD [∵ Alternate interior angles are equal]
∴ ∇ BEF ≅ Δ CED [Angle side angle congruence]
∴ BF = CD [Corresponding Parts of Congruent Triangles]
AF = AB + AF
AF = AB + AB
AF = 2 AB
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Remark: Figure is incorrect in Question
