Question

State the whether given statement is true or false

The diagonals of a parallelogram ABCD intersect at a point O. Through O, a Line is drawn to intersect AD at P and BC at Q. show that PQ divides the Parallelogram into two parts of equal area. [Hint: $\Delta AOP\cong \Delta COQ$]

• A. True
• B. False

Answer 1

The correct option is A True

In $\Delta AOB$ & $\Delta COD$

$\angle DCO=\angle BAO\left(AB||CD\right)$

$AB=CD\left(11gm\right)$

$\angle CDO=\angle ABO\left(AB||CD\right)$

$\Delta AOB\cong \Delta COD\left(ASA\right)$

Area $\left(\Delta AOB\right)=$ Are $\left(\Delta COD\right)$

In $\Delta AOP$ and $\Delta COQ$

$\angle PAO=\angle QCO\left(AD||BC\right)$

$AO=CO$ ($O$ is mid points of $AC$)

$\angle POA=\angle QOC$ (vertically opposite angles)

$\Delta AOP\cong \Delta COQ$ (ASA)

Area $\left(\Delta AOP\right)=Area\left(\Delta COQ\right)$

Similarlly,

Area $\left(\Delta DOP\right)=Area\left(\Delta BOQ\right)$

Area below $PQ$ line $=Area\left(\Delta AOB\right)+Area\left(\Delta AOP\right)+Area\left(\Delta BOQ\right)$

$=Area\left(\Delta COD\right)+Area\left(\Delta COQ\right)+Area\left(\Delta DOP\right)$

Area above $PQ$ line $=Area\left(\Delta COD\right)+Area\left(\Delta COQ\right)+Area\left(\Delta DOP\right)$

Hence $PQ$ divides $11gm$ into two parts of equal areas.