Solution:
$AC = 3\, cm$
Step by step solution:
We have $AD$ is the bisector of angle $A$
Apply Angle Bisector Theorem, we get:
$\dfrac{AC}{AB} = \dfrac{DC}{DB} = \dfrac{1,5}{2,5} = \dfrac{3}{5}$
$\Leftrightarrow AB = \dfrac{5}{3}AC$
Apply Pythagorean Theorem in $ΔABC$ with $\widehat{C}=90^\circ$ we get:
$AB^2 = AC^2 + BC^2$
$\Leftrightarrow \left(\dfrac{5}{3}AC\right)^2 = AC^2 + (1,5 + 2,5)^2$
$\Leftrightarrow \dfrac{16}{9}AC^2 = 16$
$\Leftrightarrow AC^2 = 9$
$\Leftrightarrow AC = 3\, (cm)$
