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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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In triangles BMC and DME,

`triangleBMM cong triangleEMD` ( ASA criterion)
BC=ED (corresponding parts of congruent triangle)….(1)
BC=AD (opposite sides of a prallelogram) ..(2)
Adding (1) and (2) , we get
`2BC=DE+AD Rightarrow2BC= AE`
` Rightarrow (BC)/(AE)=1/2` ....(3)
Now in `triangle BCL and triangleEAL`
`angle5=angle6 ` (alternate `angles` as BC||AE)
`angle7= angle8` ( vertically opposite angles)
`triangleBCL~ triangleEAL` (AA corollary)
` (BC)/(EA)=(BL)/(EL)` (corresponding sides of similar triangles are proportional) ...(4)
`1/2 =(BL)/(EL)` [ from (3) and (4) ]
EL=2BL Hence proved.
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