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`

Explanation

In `triangle {BMC}` and `triangle {EMD}`


`angle {BMC}=angle {EMD}` [Vertically opposite]


`{MC}={DM}` [Given]


`angle {BCM}=angle {EDM}` [Alternate angles]


`therefore triangle {BMC} cong triangle {EMD}[{By} {ASA}]`


Hence, `{BC}={DE}[{By} {CPCT}] rightarrow(1)`


`{AE}={AD}+{DE}={BC}+{BC}=2 {BC} rightarrow(2)`


Now, `triangle {BLC} sim triangle {ELA}({AA}` Similarly)


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