MCQ
If $E$ is a point on side $CA$ of an equilateral triangle $\text{ABC}$ such that $\text{BE}\perp\text{CA},$ then $AB^2+ BC^2+ CA^2=$
- A$2BE^2$
- B$3BE^2$
- ✓$4BE^2$
- D$6BE^2$
✓
Answer
Correct option: C.
$4BE^2$
In triangle $\text{ABC}, E$ is a point on $AC$ such that $\text{BE}\perp\text{AC}.$
We need to find $AB^2+ BC^2+ AC^2$
Since $\text{BE}\perp\text{AC},$
$\text{CE} = \text{AE} = \frac{\text{AC}}{2}\ $
$($In a equilateral triangle, the perpendicular from the vertex bisects the base$.)$
In triangle $\text{ABE},$ we have,
$A B^2=B E^2+A E^2$
Since $AB = BC = AC$
Therefore, $A B^2=B C^2=A C^2=B E^2+A E^2$
$\Rightarrow A B^2+B C^2+A C^2=3 B E^2+3 A E^2$
Since in triangle $BE$ is an altitude, so $\text{BE}=\frac{\sqrt{3}}{2}\text{AB}$
$\text{BE}=\frac{\sqrt{3}}{2}\text{AB}$
$=\frac{\sqrt{3}}{2}\times\text{AC}$
$=\frac{\sqrt{3}}{2}\times2\text{AE}=\sqrt{3}\text{AE}$
$\Rightarrow\text{AB}^2+\text{BC}^2+\text{AC}^2=3\text{BE}^2+3\Big(\frac{\text{BE}}{\sqrt{3}}\Big)^2$
$=3\text{BE}^2+\text{BE}^2=4\text{BE}^2$
Hence option $C$ is correct.
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