Show that a diagonal divides a parallelogram into two triangles of equal area.



Given: A parallelogram ABCD with a diagonal BD


To prove: area(∆ABD) = area(∆BCD)


Proof:


We know that in a parallelogram opposite sides are equal, that is


AD = BC and AB = CD


Now, consider ∆ABD and ∆BCD


Here AD = BC


AB = CD


BD = BD (common)


Hence by SSS congruency


∆ABD ∆BCD


By this we can conclude that both the triangles are equal


area(∆ABD) = area(∆BCD)


Hence proved


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