Question

Through the mid-point M of side DC of a rectangle ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2BL.

Answer 1

In△DME and △CMB

$DM = CM$ [∵ M is the mid-point of DC]

$∠DME = ∠CMB$ [Vertically Opposite Angles]

$∠EDM = ∠BCM$ [Alternate Angles]

∴△DME ≅△CMB [AAS Congruency]

∴ $DE = BC$ (CPCT)

$AD = BC$ [Opposite sides of Parallelogram]

Now,

$AE = AD + DE$

$= BC + DE$ [∵AD = BC]

$= 2DE$ [∵BC = DE]

Now,

In△ALE and △CLB

$∠ALE = ∠CLB$ [Vertically Opposite Angles]

$∠EAL = ∠BCL$ [Alternate Angles]

△ALE ~△CLB [By AA Similarity]

$\frac{AL}{CL} = \frac{EL}{BL} = \frac{AE}{CB}$

$\frac{AE}{CB} = \frac{EL}{BL}$

$\frac{2DE}{DE} = \frac{EL}{BL}$(DE = BC)

$EL = 2BL$

Hence proved.

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