In triangle ABC the internal bisector of ∠A meets BC at D. The measurements are AB = 12 cm, AC = 15 cm, BD = 8 cm and DC = 10 cm. What is the length of AD?

First check the data is consistent: the angle-bisector theorem needs $BD:DC=AB:AC$, and 8 : 10 = 12 : 15 = 4 : 5, so it is. The length of an internal bisector is given by $A{D}^2=AB\times AC-BD\times DC=12\times 15-8\times 10=180-80=100$, so $AD=10$ cm. (Coordinates confirm it: with $B(0,0),C$(18, 0), the vertex A falls at (6.75, √98.4375) and the distance from A to $D$(8, 0) is exactly 10.) The trap 6√5 = √180 drops the subtraction of $BD\times DC$, 13.5 averages the two sides, and 18 cm is simply $BC$.