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Through the mid-point M of the side C D ...

Through the mid-point `M` of the side `C D` of a parallelogram `A B C D` , the line `B M` is drawn intersecting `A C` at `La n dA D` produced at `E` . Prove that `E L=2B Ldot`

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`A ||gm ABCD and M ` is the midpoint of CD. Line MB is drawn, intersecting AC in L and AD production in E.
TO PROVE EL=2BL
PROOF In `Delta BMC and DeltaEMD`, we have
`angle 1 angle2` (vert opp.` angle`)
`MC=MD` [ M is the midpoint of CD]
`angle BCM= angle EDM` [ alternate interior `angle`]
`:. angle MBC~= Delta EMD`
`:. BC=DE`
But, `BC=AD` [ opposite sides of a ||gm]
`:. BC=AD=DE rArr AE= (AD+DE)= 2BC" "....(i)`
Now, in `Delta AEL and Delta CBL`, we have
`angle 6= angle5 ` (vert. opp. `angle`]
`angle 3=angle4` [ alternate interior `angle`]
`:. Delta AEl~ Delta CBL` [ AA- similarity]
`rArr (EL)/(BL)=(AE)/(BC)=(2BC)/(BC)=2` [ using (i)]
`rArr EL= 2BL`.
Hence, EL= 2BL.
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