In a $Δ A B C$ , D is midpoint of BC; AD is produced upto B so that DE =AD
i)Prove that $Δ A B D ≅ Δ B C D$
ii) AB =EC
iii) AB||EC

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Solution

In $Δ A B D$ & $Δ E C D$
i) BD =CD (Given)
$∠ A D B = ∠ E D C$ (Vertically opposite angles)
AD =ED (Given)
$∴$ $Δ A B D ≅ Δ E C D$ are concruent by SAS condition.
ii) AB =EC (CPCT)
iii) $∠ B A D = ∠ D E C (C P C T)$

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