We are given a right-angled triangle $\triangle ABC$, where $\angle B = 90^\circ$. Points $D$ and $E$ are located on side $BC$, trisecting it, with $D$ near $B$ and $E$ near $C$. We are asked to find the value of:

$\frac{8AE^2 - 3AC^2}{AD^2}$

Step-by-Step Solution:

1. Assign coordinates to the points:

   • Since $\triangle ABC$ is a right-angled triangle at $B$, we can assume that $B = (0, 0)$, $A = (0, b)$, and $C = (c, 0)$, where $b = AB$ and $c = BC$.

2. Trisection of $BC$:

   • Since $D$ and $E$ trisect side $BC$, the coordinates of these points can be computed.
   • The length of side $BC = c$, and dividing it into three equal parts:
     • $D = \left( \frac{2c}{3}, 0 \right)$
     • $E = \left( \frac{c}{3}, 0 \right)$

3. Find lengths $AE$ and $AD$:

   • Using the distance formula, $AE$ is given by: $AE = \sqrt{ \left( 0 - \frac{c}{3} \right)^2 + \left( b - 0 \right)^2 } = \sqrt{ \frac{c^2}{9} + b^2 }$
   • Similarly, $AD$ is: $AD = \sqrt{ \left( 0 - \frac{2c}{3} \right)^2 + \left( b - 0 \right)^2 } = \sqrt{ \frac{4c^2}{9} + b^2 }$

4. Substitute these into the given expression:

   • Now, calculate $AE^2$ and $AD^2$: $AE^2 = \frac{c^2}{9} + b^2, \quad AD^2 = \frac{4c^2}{9} + b^2$
   • Substitute these into the expression $\frac{8AE^2 - 3AC^2}{AD^2}$: $\frac{8AE^2 - 3AC^2}{AD^2} = \frac{8 \left( \frac{c^2}{9} + b^2 \right) - 3c^2}{\frac{4c^2}{9} + b^2 }$ Simplifying the numerator: [ 8AE^2 - 3AC^2 = 8 \left( \frac{c^2}{9} + b^2 \right) - 3c^2 = \frac{8c^2}{9} + 8b^2 - 3c^2 = \frac{-19c^2}{9} + 8b^ solution