Question : Let $\text{AX}\perp \text{BC}$ of an equilateral triangle $\text{ABC}$. Then the sum of the perpendicular distances of the sides of $\triangle \text{ABC}$ from any point inside the triangle is:
Option 1: Equal to $BC$
Option 2: Equal to $AX$
Option 3: Less than $AX$
Option 4: Greater than $AX$
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Correct Answer: Equal to $AX$
Solution : Given: $\text{AX}\perp \text{BC}$ of an equilateral triangle $\text{ABC}$. Let $\text{O}$ be a point inside the triangle. Such that $\text{OD} \perp \text{BC}$, $\text{OE} \perp \text{AC}$ and $\text{OF} \perp \text{AB}$. Since $\text{ABC}$ is an equilateral triangle. Such that $\text{AB = BC = CA}$. Area of $\triangle \text{ABC}$ = Area of $\triangle \text{OAB}$ + Area of $\triangle \text{OBC}$ + Area of $\triangle \text{OAC}$, $⇒\frac{1}{2} \times BC \times AX=\frac{1}{2} \times AB \times OF + \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AC \times OE$ $⇒\text{AX = OF + OD + OE}$ [$\because\text{AB = BC = CA}$] Hence, the correct answer is 'Equal to $AX$'.
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