jmc

We extend $AD$ to the point $E$ where it intersects the perpendicular to $BC$ from $C.$



By the Pythagorean Theorem in $△ADB,$ $B{D}^2=B{A}^2-A{D}^2=13^2-5^2=144,$ so $BD=12\text{ cm}.$

By the Pythagorean Theorem in $△DBC,$ $B{C}^2=D{C}^2-B{D}^2=20^2-12^2=256,$ so $BC=16\text{ cm}.$

Since $BCED$ has three right angles (and in fact, a fourth right angle at $E$), then it is a rectangle, so $DE=BC=16\text{ cm}$ and $CE=BD=12\text{ cm}.$

Therefore, if we look at $△AEC,$ we see that $AE=16+5=21\text{ cm},$ so by the Pythagorean Theorem, $A{C}^2=21^2+12^2=585,$ so $AC\approx \boxed{24.2}\text{ cm},$ to the nearest tenth of a centimeter.