Question

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

Answer 1

Answer:

Please see the attachment

Answer 2

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

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