Let $AC=y$. We note that $AD=AC-CD,$ so we need to find $AC$ and $CD$ separately. By Pythagoras on triangle $ABC,$ $AC=\sqrt{25^2-10^2}=15\sqrt{3}.$ Next, notice that $\triangle CDA$ and $\triangle BDA$ share hypotenuse $AD.$ Using the $HL$ congruency criterion, it follows that $\triangle CAD\cong \triangle BAD.$ In particular, $CD=DA=AC-CD.$ We then solve for $CD$ as follows.
\begin{align*} CD&=DA=AC-CD \\ \Rightarrow CD+CD&=15\sqrt{3} \\ \Rightarrow 2\cdot CD&=15\sqrt{3} \\ \Rightarrow CD&=\frac{15\sqrt{3}}{2}. \\ \end{align*}
Finally, \[AD=AC-CD=15\sqrt{3}-\frac{15\sqrt{3}}{2}=\frac{15\sqrt{3}}{2}.\]$\boxed{\frac{15\sqrt{3}}{2}}$
In right triangle $ABC$, $BD=CD+9$. If $AB=10$ and $BC=25$, what is $AD$
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