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Question
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E. Prove that: EL = 2BL.
Solution
∠1 = ∠6 (Alternate interior angles)
∠2 = ∠3 (Vertically opposite angles)
DM = MC (M is the mid-point of CD)
ΔDEM ≅ ΔCBM (AAS Congruence criterion)
So, DE = BC (Corresponding parts of congruent triangles)
Also, AD = BC (Opposite sides of a parallelogram)
AE = AD + DE = 2BC
Now, ∠1 = ∠6 and ∠4 = ∠5
ΔELA ∼ ΔBLC (AA similarity)
`=> (EL)/(BL) = (EA)/(BC)`
`=> (EL)/(BL) = (2BC)/(BC) = 2`
`=>` EL = 2BL
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