ABCD is a parallelogram and E is MID POINT OF BC .DE AND AB ON PRODUCING MEET AT F PROOF THAT AF = 2 AB
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ABCD is a parallelogram. E is the midpoint of BC. So, BE = CE.
DE produced meets the AB produced at F.
Consider the triangles CDE and BFE.
BE = CE [Given]
∠CED = ∠BEF [Vertically opposite angles]
∠DCE = ∠FBE [Alternate angles]
∴ ΔCDE ≅ ΔBFE
So, CD = BF [CPCT]
But, CD = AB
Therefore, AB = BF
AF = AB + BF
AF = AB + AB
AF = 2AB
Hence, proved
DE produced meets the AB produced at F.
Consider the triangles CDE and BFE.
BE = CE [Given]
∠CED = ∠BEF [Vertically opposite angles]
∠DCE = ∠FBE [Alternate angles]
∴ ΔCDE ≅ ΔBFE
So, CD = BF [CPCT]
But, CD = AB
Therefore, AB = BF
AF = AB + BF
AF = AB + AB
AF = 2AB
Hence, proved
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