help!

ABC has a right angle at B. \(AC = 60\), \(AE = 52\) and \(CD=39\).
Find the length of \(DE\).

\(\text{Let $AD=t$ } \\ \text{Let $BD=w$ } \\ \text{Let $BE=u$ } \\ \text{Let $CE=v$ } \\ \text{Let $DE=\color{red}x$ }\)

\(\begin{array}{|lrcll|} \hline (1): & 60^2 &=& (u+v)^2+(t+w)^2 \\ (2): & 52^2 &=& u^2+(t+w)^2 \\ (3): & 39^2 &=& (u+v)^2+w^2 \\ (4): & x^2 &=& u^2+w^2 \\ \hline (4)+(1)-(2)-(3): \\ &x^2+60^2-52^2-39^2 &=& u^2+w^2 \\ &&& +(u+v)^2+(t+w)^2 \\ &&& - \left(u^2+(t+w)^2\right) \\ &&& -\left((u+v)^2+w^2 \right) \\\\ &x^2+60^2-52^2-39^2 &=& u^2+w^2 \\ &&& +(u+v)^2+(t+w)^2 \\ &&& - u^2-(t+w)^2 \\ &&& -(u+v)^2-w^2 \\\\ &x^2+60^2-52^2-39^2 &=& 0 \\ &x^2 &=& 52^2+39^2-60^2 \\ &x^2 &=& 625 \\ &\mathbf{x} &=& \mathbf{25} \\ \hline \end{array}\)

The length of DE is \(\mathbf{25}\)