Question

prove that the diagonals of a parallelogram divide it into four triangles of equal area

Answer 1

Prove that the diagonals of a parallelogram divide it into four triangles of equal area into   two triangles of equal areas. Area (ΔAOB) = Area (ΔBOC) ... (1) In ΔBCD, CO is the median. Area (ΔBOC) = Area (ΔCOD) ... (2) Similarly, 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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