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