Question
Show that a diagonal divides a parallelogram into two triangles of equal area.
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Answer
Let ABCD be a parallelogram and BD be its diagonal.
To prove:
$\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{CDB})$
Proof:
In $\triangle\text{ABD}$ and $\triangle\text{CDB},$ we have :
AB = CD [Opposite sides of a parallelogram]
AD = CB [Opposite sides of a parallelogram]
BD = DB [Common]
i. e., $\triangle\text{ABD}\cong\triangle\text{CDB}$ [SSS criteria]
$\therefore\ \text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{CDB})$
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