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