In an equilateral triangle ABC, BC is trisected at D. Which of the following relations is correct ?

In $\triangle A B C, D$ is a point on $B C$ where $B D=\frac{1}{3} B C$, $\mathrm{AE} \perp \mathrm{BC}$ In $\triangle \mathrm{AEB}$ and $\triangle \mathrm{AEC}$, $\mathrm{AB}=\mathrm{AC}$ $\angle \mathrm{AEB}=\angle \mathrm{AEC}=90^{\circ}$ and $\mathrm{AE}=\mathrm{AE}$ By RHS-criterion $\triangle \mathrm{AEB} \sim \triangle \mathrm{AEC}$ $\Rightarrow \mathrm{BE}=\mathrm{EC}$ $\therefore \mathrm{BD}=\frac{1}{3} \mathrm{BC}, \mathrm{DC}=\frac{2}{3} \mathrm{BC}$ and $\mathrm{BE}=\mathrm{EC}=\frac{1}{2} \mathrm{BC} \ldots .$ (i) $\angle \mathrm{C}=60^{\circ}$ $\therefore \triangle \mathrm{ADC}$ is an acute angled triangle. $\therefore \mathrm{AD}^{2}=\mathrm{AC}^{2}+\mathrm{DC}^{2}-2 \mathrm{DC} \times \mathrm{EC}$ $\Rightarrow \mathrm{AD}^{2}=\mathrm{AC}^{2}+\left(\frac{2}{3} \mathrm{BC}\right)^{2}-2 \times \frac{2}{3} \mathrm{BC} \times \frac{1}{2} \mathrm{BC}$ $\Rightarrow \mathrm{AD}^{2}=\mathrm{AC}^{2}+\frac{4}{9} \mathrm{BC}^{2}-\frac{2}{3} \mathrm{BC}^{2}$ $\Rightarrow \mathrm{AD}^{2}=\mathrm{AB}^{2}+\frac{4}{9} \mathrm{AB}^{2}-\frac{2}{3} \mathrm{AB}^{2}$ $[\because \mathrm{AB}=\mathrm{BC}=\mathrm{CA}]$ $\Rightarrow \mathrm{AD}^{2}=\frac{9 \mathrm{AB}^{2}+4 \mathrm{AB}^{2}-6 \mathrm{AB}^{2}}{9}=\frac{7}{9} \mathrm{AB}^{2}$ $\Rightarrow 9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2}$ Alternative:

Let,

$\mathrm{A} \mathrm{B}=\mathrm{BC}=\mathrm{AC}=a$ (in equilateral triangle, sides are equal)

$\mathrm{BD}=\frac{a}{3}$

$\mathrm{DE}=\left(\frac{a}{2}-\frac{a}{3}\right)=\frac{a}{6}$

$\because$ Height of equilateral triangle $=\frac{\sqrt{3} a}{2}$

By Pythagoras theorem

(Hypotenuse)² = (Perpendicular)² = (Base)²

AD² = AE² + DE²

$=\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{a}{6}\right)^{2}$

$=\frac{3}{4} a^{2}+\frac{a^{2}}{36}$

$=\frac{27 a^{2}+a^{2}}{36}$

$\mathrm{AD}^{2}=\frac{28 a^{2}}{36}$

9AD² = 7a² or 9AD² = 7AB²