In the adjoining figure, AB = CD and AB||CD prove that
(i) `DeltaAOB cong DeltaDOC`
(ii) AD and BC bisect each other at the point O.

Explanation

(i) since AB||CD (given)
`implies " "angle1=angle2` and `angle3=angle4` (alternate angles)
Now, in `DeltaAOB` and `DeltaDOC`,
`:' {(angle1=angle2,"(alternate angles)"),(AB=DC,"(given)"),(angle3=angle4,"(alternate angle)"):}`
Therefore, `DeltaAOB cong DeltaDOC` (ASA) Hence Proved.
(ii) Therefore, `AO=DO` (c.p.c.t.)
`implies BC` bisects AD.
Also `BO = CO` (c.p.c.t)
`implies AD` bisects BC.
Hence, AD and BC bisect each other. Hence Proved.