Question

a line ab is drawn through the point of intersection of the diagonals of a parallelogram as shown in the figure prove that triangle ROA is congruent to triangle POB give reason for your answer

Answer 1

In ∆ROA and ∆POB

<ROA = <POB. [Vertically opposite angles]

<ARO = <BPO. [Alternate interior angles]

PO = RO.
[Diagonals of parallelogram bisect each other]

∆ROA =~ ∆POB. [By ASA]

Answer 2

Triangle POB and triangle ROA are congruent.

Step-by-step explanation:

1. SR is parallel to PQ because PQRS is a parallelogram.

 SR ll PQ     ...1)

  We can also say

  AR ll BP      ...2)

2. AB is a transverse line for parallel line AR and BP. Then alternate interior angle should be equal.

   $\angle RAO=\angle PBO$         ...3)

3. Now $\Delta ROA$ and $\Delta BOP$

    $\angle RAO=\angle PBO$           ( alternate interior angle)

   

     $\angle ROA=\angle POB$           ( Vertical opposite angle)

 

    OR = OP         ( because diagonal of parallelogram bisect each other)

 

   So

   $\Delta ROA \cong \Delta POB$      (by AAS method)