Question
In a parallelogram $\text{ABCD},$ any point $E$ is taken on the side $\text{BC. AE}$ and $DC$ when produced meet at a point $M.$ Prove that $\text{ar}(\triangle\text{ADM})=$ ar $(\text{ABMC}).$
✓
Answer
Construction: Join $AC$ and $BM$
Let $h$ be the distance between $AB$ and $CD$.
$\text{A}(\triangle\text{ACD})=\frac{1}{2}\times\text{CD}\times\text{h}$
$\text{A}(\triangle\text{ABM})=\frac{1}{2}\times\text{AB}\times\text{h}$
$=\frac{1}{2}\times\text{CD}\times\text{h} [AB = CD,$ opposite sides of $\|^m \text{ABCD}]$
$\Rightarrow\ \text{A}(\triangle\text{ABM})=\text{A}(\triangle\text{ACD})$
$\Rightarrow\ \text{A}(\triangle\text{ABM})+\text{A}(\triangle\text{ACM})=\text{A}(\triangle\text{ACD})+\text{A}(\triangle\text{ACM})$
$\Rightarrow\ \text{A}(\text{ABMC})=\text{A}(\triangle\text{ADM})$
Let $h$ be the distance between $AB$ and $CD$.
$\text{A}(\triangle\text{ACD})=\frac{1}{2}\times\text{CD}\times\text{h}$
$\text{A}(\triangle\text{ABM})=\frac{1}{2}\times\text{AB}\times\text{h}$
$=\frac{1}{2}\times\text{CD}\times\text{h} [AB = CD,$ opposite sides of $\|^m \text{ABCD}]$
$\Rightarrow\ \text{A}(\triangle\text{ABM})=\text{A}(\triangle\text{ACD})$
$\Rightarrow\ \text{A}(\triangle\text{ABM})+\text{A}(\triangle\text{ACM})=\text{A}(\triangle\text{ACD})+\text{A}(\triangle\text{ACM})$
$\Rightarrow\ \text{A}(\text{ABMC})=\text{A}(\triangle\text{ADM})$
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