# <span lang="EN-US

Given : A parallelogram ABCD with a point ‘O’ inside it.

To prove : (i) area(∆OAB) + area(∆OCD) = area(gm ABCD),

(ii)area(∆OAD) + area(∆OBC) = area(gm ABCD).

Construction : Draw PQ AB and RS AD

Proof:

(i)

∆AOB and parallelogram ABQP have same base AB and lie between parallel lines AB and PQ.

According to theorem: If a triangle and parallelogram are on the same base and between the same

parallel lines, then the area of the triangle is equal to half of the area of the parallelogram.

area(∆AOB) = area(gm ABQP) –--1

Similarly, we can prove that area(∆COD) = area(gm PQCD) –--2

Adding –1 and –2 we get,

area(∆AOB) + area(∆COD) = area(gm ABQP) + area(gm PQCD)

area(∆AOB) + area(∆COD) = = area(gm ABCD)

area(∆AOB) + area(∆COD) = area(gm ABCD)

Hence proved

(ii)

According to theorem: If a triangle and parallelogram are on the same base and between the same

parallel lines, then the area of the triangle is equal to half of the area of the parallelogram.

Similarly, we can prove that area(∆OBC) = area(gm BCRS) –--2

Adding –1 and –2 we get,

area(∆OAD) + area(∆OBC) = area(gm ASRD) + area(gm BCRS)

area(∆OAD) + area(∆OBC) = = area(gm ABCD)

area(∆OAD) + area(∆OBC) = area(gm ABCD)

Hence proved

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