Question

In the adjoining figure, line segment AB is parallel to another line segment CD. O is mid-point of AD. Show that
(i) ∆AOB ≅ ∆DOC
(ii) O is also mid-point of BC.

Answer 1

ΔAOB ≅ ΔDOC by ASA congruency.

AB = CD (Given)

O is the mid - point (Given)

AB || CD and BC is the transversal

In ΔAOB and ΔDOC ,

AB = CD ( given )

∠OAB = ∠ODC ( alternate interior angles )

∠AOB = ∠DOC ( vertically opposite angles )

Therefore , ΔAOB ≅ ΔDOC ( ByASA rule )

Thus,

OB = OC ( By CPCT )

Hence , O is also the mid point of BC