jmc

Let $h=AD$. Then by Pythagoras on right triangle $ABD$, $$B{D}^2=10^2-h^2=100-h^2,$$and by Pythagoras on right triangle $ACD$, $$C{D}^2=17^2-h^2=289-h^2.$$

But $BD:CD=2:5$, so $B{D}^2:C{D}^2=4:25$. Hence, $$\frac{100-h^2}{289-h^2}=\frac{4}{25}.$$Solving for $h$, we find $h=\boxed{8}$.