Let \(ABCD\) be a cyclic quadrilateral with \(AB = 15, BC = 20, CD = 24\) and \(AC = 25\). Then \(AD\) equals __________

The length $AD$ is $7$. Since $15^2+20^2=25^2$, angle $ABC$ is a right angle and $AC$ is a diameter of the circumcircle. Then angle $ADC$ is also a right angle, so $AD^2=25^2-24^2=49$. The value $49$ is the square of the requested length.