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# Diagonals ac and bd of a trapezium abcd with ab parallel to CD intersect each other at o.prove that area of triangle aod equals area of boc?

Updated: 10/19/2022

Wiki User

6y ago

First consider triangles ACD and BCD.
They share a common base, CD, and their height is the distance between the parallel lines AB and CD.
Consequently, area(ACD) = area(BCD) . . . . . . . . eqn 1
But ACD = AOD + OCD and BCD = BOC + OCD
So area(ACD) = area(AOD) + area(OCD)
and area(BCD) = area(BOC) + area(OCD)

Substituting in eqn 1,
area(AOD) + area(OCD) = area(BOC) + area(OCD)
area(AOD) = area(BOC)

Wiki User

6y ago

Wiki User

13y ago

First consider triangles ACD and BCD.

They share a commom base, CD, and their height is the ditance between the parallel lines AB and CD.

Consequently, area(ACD) = area(BCD) . . . . . . . . eqn 1

But ACD = AOD + OCD and BCD = BOC + OCD

So area(ACD) = area(AOD) + area(OCD)

and area(BCD) = area(BOC) + area(OCD)

Substituting in eqn 1,

area(AOD) + area(OCD) = area(BOC) + area(OCD)

area(AOD) = area(BOC)