Question

ABCD is a parallelogram whose diagonals intersect
at O. If P is any point on BO, prove that:
(i) ar (ADO) = ar (CDO)
(ii) ar (ABP) = ar (CBP).​

Answer 1

Answer:

arADO=AarCDO

Step-by-step explanation:

Since the diogonal bisect each other

O IS MID POINT OF AC AND BD

OF IS MEDIAN

AR ADO= AR CDO

Answer 2

Step-by-step explanation:

Given : ABCD is a parallelogram whose diagonals intersect at O and P is any point on BO.  

 

 

To Prove : (i) ar(Δ ADO) = ar(Δ CDO)

(ii) ar(Δ ABP) = ar(Δ CBP)

 

Proof :  

Since diagonals of a parallelogram bisect each other. Therefore, O is the midpoint of AC as well as BD.

AO = OC and BO = OD

 

(i) In ΔACD, DO is a median.

Therefore,

ar (ΔADO) = ar (ΔCDO)

Hence, proved

 

(ii) since O is the midpoint of AC. therefore , OP & OB are medians of ∆APC & ∆ABC.

In ∆APC, since OP is the median.Then,

ar (ΔAOP) = ar (ΔCOP) ……...(i)

In ∆ABC , since OB is a median.Then,

ar (ΔAOB) = ar (ΔCOB) ………...(ii)

On Subtracting eq(i) from (ii), we get :  

ar (ΔAOB) - ar (ΔAOP) = ar (ΔCOB) - ar (ΔCOP)

ar (ΔABP) = ar (ΔCBP)

Hence, proved