Equilateral triangle $ABC$ with $D$ on $BC$ such that $BD=\frac13BC$. Prove $9\overline{AD}^2=7\overline{AB}^2$

Question

Equilateral triangle $ABC$, point $D$ on side $BC$ such that $BD=\frac13BC$. Prove $9\overline{AD}^2=7\overline{AB}^2$

In an equilateral triangle $ABC$, $D$ is a point on side $BC$ such that $BD=\frac{1}{3}BC$. Prove that $9\overline{AD}^2=7\overline{AB}^2$.

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I started by constructing the median/altitude(because it's an equilateral triangle so it is the same thing) $AE$. $$BE=\frac{BC}{2}$$ $$DE=\frac{BC}{2}-\frac{BC}{3}=\frac{BC}{6}$$ $$\overline{AD}^2=\overline{AE}^2+\frac{\overline{BC}^2}{36}$$ $$\overline{AB}^2=\overline{AE}^2+\frac{\overline{BC}^2}{4}$$ But I got $$9\overline{AD}^2=\overline{AB}^2$$ Am I making a mistake somewhere or is there a problem in the question?

Answer 1

You had an error. It should be $\overline{BE}^2=\frac14\overline{AB}^2$and

$$\overline{AD}^2=\overline{AE}^2+\frac{\overline{BC}^2}{36} =\overline{AB}^2- \frac14 \overline{AB}^2+\frac{\overline{AB}^2}{36} = \frac79\overline{AB}^2$$

Answer 2

It is enough to apply Stewart's theorem. Assuming $AB=AC=BC=3$ and $AD=DE=EC=1$, then letting $AD=x$, we have $\cos\widehat{ADB}+\cos\widehat{CDA}=0$, hence

$$ \frac{1+x^2-9}{2x}+\frac{4+x^2-9}{4x}=0 $$ and $x=\sqrt{7}$, such that $AD^2=\frac{7}{9}AB^2$ as wanted. As a simpler alternative, if we denote as $M$ the midpoint of $DE$ we have $DM=\frac{1}{6}AB,\,AM=\frac{\sqrt{3}}{2}AB$ and the conclusion follows from the Pythagorean theorem.