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

Solution:

O is the mid point of AC and BD. (diagonals of bisect each other)

In ΔABC, BO is the median.

∴ar(AOB) = ar(BOC) — (i)

also,

In ΔBCD, CO is the median.

∴ar(BOC) = ar(COD) — (ii)

In ΔACD, OD is the median.

∴ar(AOD) = ar(COD) — (iii)

In ΔABD, AO is the median.

∴ar(AOD) = ar(AOB) — (iv)

From equations (i), (ii), (iii) and (iv), we get,

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

Hence, we get, the diagonals of a parallelogram divide it into four triangles of equal area.