Q. 204.5( 26 Votes )

Through the midpoint

Answer :

Given: ABCD is a parallelogram, M is the midpoint of CD. BM intersects AC at L and AD produced at E.
To prove: EL = 2BL
Proof: In Δ BMC and Δ EDM
∠DME = ∠ BMC (Vetically opposite angles)
DM = MC (given)
∠DEM = ∠ MBC (alternate angles)
∴ Δ BMC  ≅ Δ EDM (ASA congruence)
∴ DE = BC (c.p.c.t)
But BC = AD (opposite sides of parallelogram ABCD)
∴ AD = DE ⇒ AE = 2AD = 2BC
In Δ AEL and Δ CBL
∠ ALE = ∠ BLC (Vertically opposite angles)
∠ AEL = ∠ LBC (alternate angles)
∴ Δ AEL ~ Δ CBL (AA similarity axiom)
∴ EL = 2 BL

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