Geometry

First off, let's let $DC = 3BD$

We can write two equations off the problem

$AD = \sqrt{AC^2 - (3BD)^2 } = \sqrt{15^2 - 9BD^2 } \\ AD = \sqrt{AB^2 - BD^2} = \sqrt{10^2 - BD^2 }$

Notice that we have AD on both equations, so we can equate the two to get

$\sqrt{15^2 - 9BD^2} = \sqrt {10^2 -BD^2 }$

Now, we just have to solve for BD^2. 

$15^2 - 9BD^2 = 10^2 - BD^2 \\ 225 - 9BD^2 = 100 - BD^2 \\ 125 = 8BD^2 \\ 125/8 = BD^2$

We know that 

$AD =\sqrt{10^2 - BD^2}$ so we can just plug in BD^2 to find AD. 

$AD = \sqrt{100 - 125/8} = \sqrt { 675/8} = \frac{15\sqrt3 }{2 \sqrt2}$

This rounds to about 9.19. 

Thanks! :)