Trisect Side of Equilateral Triangle

Let the side length of $\triangle ABC$ be $1$ and $N$ be the leg of altitude from $A$ to $BC$ . $N$ is also the midpoint of $BC$ . Then, we have:

$\begin{aligned} \frac {\color{#3D99F6}AD^2}{AB^2} & = \frac {\color{#3D99F6}AN^2+ND^2}{1^2} & \small {\color{#3D99F6}\text{By Pythagorean theorem.}} \\ & = \left(\frac {\sqrt 3}2\right)^2 + \left(\frac 16 \right)^2 \\ & = \frac 34 + \frac 1{36} = \frac 79 \end{aligned}$

$\implies m+n = 7+9 = \boxed{16}$

WLOG, let's assume the coordinates of $A,B,C,D$ as $(\dfrac a 2, \dfrac {a \sqrt3} 2 ),(0,0), (a,0),(\dfrac a 3,0)$ respectively.

By distance formula we get $AD^2=\dfrac{7a^2}{9}$ and $AB^2=a^2$ .

Hence $\dfrac{AD^2}{AB^2}=\dfrac{7}{9}$ .

Hence $m+n=7+9=16$ .

Yes there is one more possible solution, we can apply stewart's theorem: let in equilateral $\triangle ABC$ , $AB=BC=CA = a$ now,

$\begin{aligned} (AB^2)CD + (AC^2)BD = BC(AD^2 + BD \times CD) \\ \implies a^2 \times \frac{2a}{3} + a^2 \times \frac{a}{3} = a( AD^2 + \frac{a}{3} \times \frac{2a}3) \\ \implies AD^2 = \frac{7a^2}{9} \\ \implies \frac{AD^2}{AB^2} = \huge \boxed { \frac{7}9 }\end{aligned}$

since $\color{#3D99F6} BD= \dfrac{a}3; CD= a - \dfrac{a}3 = \dfrac{2a}{3}$ . sorry for incovience of diagram..

Use area of the equilateral triangle, ratio of areas of 2 triangles = ratio of the bases of 2 triangle if they share a common height and lastly the Pythagorean theorem.

Apply stewarts theorem to directly get answer