In the adjoining figure ABCD is a parallelogram, ABM is a line segment and E is the mid-point of BC. Prove that :
(i) `DeltaDCE cong DeltaMBe` (ii) `AB = BM`
(iii) `AM=2DC`

Explanation

(i) Since ABCD is a parallelogram
`:. ABM"||"DC`
`:. angle1=angle2` (alternate angles)
Now, in `DeltaDCE` and `DeltaMBE`
`:. {(angle1=angle2,"(proved above)"),(angle3=angle4,"(vertically opposite angles)"),(CE=BE,"(given as E is the mid-point)"):}`
`:. DeltaDCE cong DeltaMBE` (AAS) Hence Proved.
(ii) Therefore, `DC = MB` (c.p.c.t.)
But `DC=AB` (opposite sides of a parallelogram are equal)
`:. AB = MB` Hence Proved.
(iii) Now, `AM=AB+BM`
`=DC+DC` (poved above)
`=2 DC` Hence Proved.