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