Join $AE$. Since $\angle ABE = 90^\circ$ and $ABED$ is a cyclic quadrilateral, then $AE$ is a diameter of the circle passes through $A$, $B$, $E$ and $D$ (converse of $\angle$ in semi-circle). In $\Delta ABE$,
$\begin{array}{rcll}
AE^2 & = & AB^2 + BE^2 & \text{(Pyth. Thm.)} \\
AE^2 & = & 660^2 +275^2 \\
AE & = & 715\text{ cm}
\end{array}$
Since $AE$ is a diameter, then $\angle ADE=90^\circ$ ($\angle$ in semi-circle). In $\Delta ADE$,
$\begin{array}{rcll}
DE^2 & = & AE^2 -AD^2 & \text{(Pyth. Thm.)} \\
DE^2 & = & 715^2 -572^2 \\
DE & = & 429\text{ cm}
\end{array}$
In $\Delta ABC$ and $\Delta EDC$,
$\begin{array}{rcll}
\angle ABC & = & \angle EDC & \text{(ext. $\angle$, cyclic quad.)} \\
\angle ACB & = & \angle ECD & \text{(common $\angle$)} \\
\end{array}$
$\therefore \Delta ABC \sim \Delta EDC$ (A.A.A.).
Hence, we have
$\begin{array}{rclll}
\dfrac{AC}{EC} & = & \dfrac{AB}{ED} & = & \dfrac{BC}{DC} & \text{(corr. sides, $\sim\Delta$)} \\
\dfrac{AD +CD}{EC} &= & \dfrac{AB}{ED} & = & \dfrac{BE +CE}{CD} \\
\dfrac{572 +CD}{EC} &= & \dfrac{660}{429} & = & \dfrac{275 +CE}{CD} \\
\dfrac{572 +CD}{EC} &= & \dfrac{20}{13} & = & \dfrac{275 +CE}{CD}
\end{array}$
Hence, we have
$\left\{ \begin{array}{ll}
\dfrac{572 +CD}{EC} = \dfrac{20}{13} & \ldots \unicode{x2460} \\
\dfrac{20}{13} = \dfrac{275 +CE}{CD} & \ldots \unicode{x2461}
\end{array}\right.$
From $\unicode{x2460}$, we have
$\begin{array}{rcl}
\dfrac{572 +CD}{EC} & = & \dfrac{20}{13} \\
EC & = & \dfrac{13(572 +CD)}{20} \ldots \unicode{x2462}
\end{array}$
Sub. $\unicode{x2462}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
\dfrac{20}{13} & = & \dfrac{1}{CD} \left(275 +\dfrac{13(572 +CD)}{20} \right) \\
\dfrac{20CD}{13} & = & \dfrac{5500 +7436 +13CD}{20} \\
400CD & = & 168168 +169CD \\
231CD & = & 168168 \\
CD & = & 728 \text{ cm}
\end{array}$
