Equilateral triangle trisected

Question

In an equilateral triangle $ABC$ the side $BC$ is trisected by points $D$ and $E$. Prove $9|AD|^2 = 7|AB|^2$.

Answer 1

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = $\frac{BC}{2}$ = $\frac{a}{2}$

And using phythagorous theorem,

AE = $\sqrt{AB^2 - BE^2}$

AE = $\frac{\sqrt{3}a}{2}$

Given that, BD = $\frac{1}{3}$ BC

∴ BD = $\frac{a}{3}$

DE = BE - BD = $\frac a{2} - \frac a{3} = \frac a{6}$

Applying Pythagoras theorem in ΔADE, we obtain

$AD^2 = AE^2 + DE^2$

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

=$\left(\frac{3a^2}{4}\right) + \left(\frac{a^2}{36}\right)$

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

$9 AD^2 = 7 AB^2$

Answer 2

Use the Law of Cosines: \begin{align}|AD|^2 &=|AB|^2 +|BD|^2-2|AB|\cdot |BD|\cos(60^{\circ})\\ &=|AB|^2\left(1+\frac{1}{3^2}-2\cdot\frac{1}{3}\cdot\frac{1}{2}\right)\\ &=\frac{7}{9}\cdot |AB|^2. \end{align}