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

Given: ABCD be a parallelogram whose diagonal intersect at O.

To Prove: (1) ar(ΔADO) = ar(ΔCDO) (2) ar(ΔABP) = ar(ΔCBP)

Proof :

AC is the diagonal of the parallelogram.

∴ AO = OC [Diagonal of a parallelogram bisect each other]

O is the midpoint of AC Þ DO is the median.

⇒ ar(ΔADO) = ar(ΔCDO) [Median of a triangle divides it into two triangles of equal area.] --------- (1)

Similarly BO is the median of ΔABC.

⇒ ar(ΔABO) = ar(ΔBCO) [Median of a triangle divides it into two triangles of equal area.] --------- (2)

P is a point on BO. Therefore, OP is also the median of ΔAPC.

⇒ ar(ΔOPA) = ar(ΔOPC) --------- (3)

Subtracting equation (3) from equation (3), we get,

ar(ΔABO) − ar(ΔOPA) = ar(ΔBCO) − ar(ΔOPC) ⇒ ar(ΔABP) = ar(ΔCBP)

Hence proved.

Answer 2

ANSWER:

ABCD is a parallelogram and AC,BD are diagonals bisect each other.

(i) Since, Diagonals of a parallelogram bisect each other.

∴ AO=OC

⇒ O is the mid-point of AC.

DO is a median of △DAC

We know that, a median of a triangles divides it into two triangles of equal areas

∴ ar(△ADO)=ar(△CDO)

We know, a median of a triangle divides it into triangles of equal areas.

(ii) Since, BO is a median of △BAC,

∴ ar(△BOA)=ar(△BOC) ---- ( 1 )

PO is a median of △PAC

We know, a median of a triangle divides it into triangles of equal areas.

∴ ar(△POA)=ar(△POC) ---- ( 2 )

Subtracting ( 2 ) from ( 1 ) we get,

⇒ ar(△BOA)−ar(△POA)=ar(△BOC)−ar(△POC)

⇒ ar(△ABP)=ar(△CBP)