Question
In the adjoining figure, $ABC$ and $ABD$ are two triangles on the same base $AB$. If line segment $CD$ is bisected by $AB$ at $O$, show that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
✓
Answer
We know that median of a triangle divides it into two triangles of equal area. Now, $AO$ is the median of $\triangle\text{ACD}.$
$\Rightarrow\ \text{A}(\triangle\text{COA})=\text{A}(\triangle\text{DOA})\dots(1)$ And, Bo is the median of $\triangle\text{BCD}.$
$\Rightarrow\ \text{A}(\triangle\text{COB})=\text{A}(\triangle\text{DOB})\dots(2)$ Adding $(1)$ and $(2),$ we get: $\text{A}(\triangle\text{COA})+\text{A}(\triangle\text{COB})=\text{A}(\triangle\text{DOA})+\text{A}(\triangle\text{DOB})$
$\Rightarrow\ \text{A}(\triangle\text{ABC})=\text{A}(\triangle\text{ABD})$
$\Rightarrow\ \text{A}(\triangle\text{COA})=\text{A}(\triangle\text{DOA})\dots(1)$ And, Bo is the median of $\triangle\text{BCD}.$
$\Rightarrow\ \text{A}(\triangle\text{COB})=\text{A}(\triangle\text{DOB})\dots(2)$ Adding $(1)$ and $(2),$ we get: $\text{A}(\triangle\text{COA})+\text{A}(\triangle\text{COB})=\text{A}(\triangle\text{DOA})+\text{A}(\triangle\text{DOB})$
$\Rightarrow\ \text{A}(\triangle\text{ABC})=\text{A}(\triangle\text{ABD})$
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