Question

Prove that the diagonals of a parallelogram divides into four triangle of an equal area

Answer 1

Consider a llgm ABCD , in which AC & BD are its diagonals. ......... ∵ Diagonals of a llgm bisect each other, ∴ AC & BD bisect at point O ( let ).............. In Δ ABC, BO is the median, ∴ ar(Δ AOB ) = ar ( Δ BOC )........ eq. 1......................... In ΔBCD, OC is the medium, ∴ ar(ΔBOC) = ar ( Δ COD)......... eq. 2................... Similarly, ar(ΔCOD)= ar (ΔAOD )................ eq. 3................... ∴ From eq. 1,2,3 ; ar(ΔAOB) = ar ( ΔBOC ) = ar ( ΔCOD) = ar ( AOD )................ ∴diagonals of a llgm divides it into four triangles of equal area...........

Answer 2

Let us consider in a parallelogram ABCD the diagonals AC and BD are cut at point O.

To prove: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)

Proof:

In parallelogram ABCD the diagonals bisect each other.

AO = OC

In ∆ACD, O is the mid-point of AC. DO is the median.

ar (∆AOD) = ar (COD) ….. (1) [Median of ∆ divides it into two triangles of equal arreas]

Similarly, in ∆ ABC

ar (∆AOB) = ar (∆COB) ….. (2)

In ∆ADB

ar (∆AOD) = ar (∆AOB) …. (3)

In ∆CDB

ar (∆COD) = ar (∆COB) …. (4)

From (1), (2), (3) and (4)

ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)

Hence proved.

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