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

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

In $$\triangle$$ ABC, BO is the median.
$$\therefore$$ ar(AOB) = ar(BOC) .... (i)
also,
In $$\triangle$$ BCD, CO is the median.
$$\therefore$$ ar(BOC) = ar(COD) ...... (ii)
In $$\triangle$$ ACD, OD is the median.
$$\therefore$$ ar(AOD) = ar(COD) ....... (iii) In $$\triangle$$ ABD, AO is the median.
$$\therefore$$ 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.