Question

2. In Fig. 13.23, AB = CD and AB || CD.
Prove that AABO # ADCO:
​

Answer 1

GIVEN

• AB= CD
• AB ll CD

To prove

$\sf{ \triangle \: ABO ≅\: \triangle \: DCO}$

Solution

consider ∆ ABO and ∆ DCO

• AB= CD [ given]

• angle BOA= angle DOC [ vertically opposite angles ] since AB ll CD

• angle OBA = OCD [ vertically opposite angles] since AB ll CD

Therefore by AAS criteria ∆ ABO ≅ DCO

--------------------------------------

Answer 2

Given : AB = CD

AB || CD

To prove: ∆ ABO ≅ ∆ DCO

AB = CD (given)

∠ABO = ∠DCO (V.O.A)

∠B = ∠C ( Alternate interior angle)

∆ ABO ≅ ∆ DCO ( by ASA congruence condition)