jmc

Let $AB=x$ and $AC=y$. Then we can write two Pythagorean equations from the information given: $(x/3{)}^2+y^2=28^2$ and $x^2+(y/3{)}^2=16^2$. These equations become $x^2/9+y^2=784$ and $x^2+y^2/9=256$. Multiplying them both by 9, we get $x^2+9y^2=7056$ and $9x^2+y^2=2304$. Now we add the two equations to get $10x^2+10y^2=9360$, which can be reduced to $x^2+y^2=936$. We do not need to solve for $x$ and $y$ since 936 is the square of the hypotenuse $BC$. The length is thus $\sqrt{936}=\sqrt{(36\times 26)}=\sqrt{36}\times \sqrt{26}=\boxed{6\sqrt{26}}$ units.