Question
If $ABCD$ is a parallelogram, then prove that $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})\\ \ =\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$ $(||^{gm} ABCD)$
✓
Answer
Given, $ABCD$ is a parallelogram.
To prove: $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})=\text{ar}(\triangle\text{ABC})\\ \ =\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$ $(||^{gm} ABCD)$
Proof: We know that diagonal of a parallelogram divides it into two equal triangles.
Since, $AC$ is the diagonal
Then, $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar} (||^{gm} ABCD) ...(1)$
Since, $BD$ is the diagonal
Then $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})=\frac{1}{2}\text{ar} (||^{gm} ABCD) ...(2)$
Compare equation $(1)$ and $(2)$
$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\text{ar}(\triangle\text{ABD})\\ \ =\text{ar}(\triangle\text{BCD})=\frac{1}{2}\text{ar}$ ($||^{gm} ABCD$)
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