If E is a point on side CA of an equilateral triangle ABC such that BE ⊥ CA,

Question:

If $E$ is a point on side $C A$ of an equilateral triangle $A B C$ such that $B E \perp C A$, then $A B^{2}+B C^{2}+C A^{2}=$

(a) $2 \mathrm{BE}^{2}$

(b) $3 \mathrm{BE}^{2}$

(c) $4 \mathrm{BE}^{2}$

(d) $6 \mathrm{BE}^{2}$

Solution:

In triangle $\mathrm{ABC}, \mathrm{E}$ is a point on $\mathrm{AC}$ such that $B E \perp A C$.

We need to find $A B^{2}+B C^{2}+A C^{2}$.

Since $B E \perp A C, C E=A E=\mathrm{AC} 2$ (In a equilateral triangle, the perpendicular from the vertex bisects the base.)

In triangle 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 $B E=\frac{\sqrt{3}}{2} A B$

$B E=\frac{\sqrt{3}}{2} A B$

$=\frac{\sqrt{3}}{2} \times A C$

$=\frac{\sqrt{3}}{2} \times 2 A E=\sqrt{3} A E$

$\Rightarrow A B^{2}+B C^{2}+A C^{2}=3 B E^{2}+3\left(\frac{B E}{\sqrt{3}}\right)^{2}$

$=3 B E^{2}+B E^{2}=4 B E^{2}$

Hence option (c) is correct.

 

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