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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