Question

In the given fig AB || CD and AB = CD ,
Prove that -: (i) ∆AOB
_
= ∆DOC (ii) AO = DO (iii) BO = CO
​

Answer 1

AB||DC

And

AB=CD

1.AOB=DOC.... Alternate angles are congruent.

2.AO=DO.... Lying on same line and are intersecting other line.

3.BO=CO...Interesting each other.

All the two sides of a triangle are congruent

Answer 2

Given, AB || CD and AB = CD

As line AB is parallel to CD, by the properties of parallel lines, $\angle$ ABO should be equal to $\angle$ OCD.

Therefore,

$\angle$ ABO = $\angle$ OCD ----: ( 1 )

Also, $\angle$ AOB is vertically opposite to $\angle$ COD, and we know that vertically opposite angles are always equal.

Therefore,

$\angle$ AOB= $\angle$ COD ----: ( 2 )

Now,

= > AB = CD ( Given in the question )

= > $\angle$ ABO = $\angle$ OCD ( From 1 )

= > $\angle$ AOB = $\angle$ COD.

Thus,

$\triangle$ AOB $\cong \triangle$ DOC, by ASA.

And, hence AO = DO,also BO = CO.

Hence proved.