Question

In quadrilateral abcd diagonal ac bisects diagonal bd at point o prove that ar(abc)=1/2ar(abcd)

Answer 1

proved that Area of Δ ABC = Area of ABCD /2 if In quadrilateral abcd diagonal ac bisects diagonal bd at point o

Step-by-step explanation:

In quadrilateral abcd diagonal ac bisects diagonal bd at point o

Now AO = OC   & BO = OD

in Δ ABC   , BO bisects AC as AO = OC

=> Area of Δ AOB = Area of Δ COB

Simialrly

Area of Δ AOD = Area of Δ COD

now in ΔABD , AO bisects BD as BO = OD

=> Area of Δ AOB = Area of Δ AOD

Hence area of all sections is Equal

Area of ABCD = Area of Δ AOB + Area of Δ COB + Area of Δ AOD + Area of Δ COD  

= 4  Area of Δ AOB   ( as  Area of Δ AOB  = Δ COB = Δ AOD = Δ COD   )

=> Area of Δ AOB  = Area of ABCD /4

Area of Δ ABC = Area of Δ AOB + Area of Δ COB

= 2 Area of Δ AOB

= 2 ( Area of ABCD /4)

= Area of ABCD /2

Area of Δ ABC = Area of ABCD /2

Learn more :

Prove that prove that the diagonals of a rectangle bisect each other

https://brainly.in/question/12831299

diagonals of a parallelogram bisected each other

https://brainly.in/question/1897091

Answer 2

Answer with Step-by-step explanation:

Diagonal AC and diagonal BD bisect at point O.

In triangle ABC, OB bisect diagonal AC

Ar(triangle AOB)=Ar(triangle BOC)

In triangle ACD, OD bisects AC

Ar (triangle AOD)=Ar(triangle COD)

In triangle ABD, OA bisects BD

Ar(AOD)=Ar(AOB)

Therefore, Ar(AOB)=Ar(AOC)=Ar(AOD)=Ar(COD)

Ar(ABCD)=Ar(AOB)+Ar(AOC)+Ar(AOD)+Ar(COD)

Substitute the values then we get

$Ar(ABCD)=Ar(AOB)+Ar(AOB)+Ar(AOB)+Ar(AOB)=4Ar(AOB)$

$Ar(AOB)=\frac{1}{4}ABCD$

$Ar(ABC)=Ar(AOB)+Ar(BOC)=Ar(AOB)+Ar(AOB)=2Ar(AOB)$

Substitute the value

Ar(ABC)=$2\times \frac{1}{4}ABCD=\frac{1}{2}ABCD$

Hence, Ar(ABC)=$\frac{1}{2}ABCD$

#Learns more:

https://brainly.in/question/8227816:Answered by Sainilavish