Let $ABCD$ be a cyclic quadrilateral with $AB = 15, BC = 20, CD = 24$ and $AC = 25$. Then $AD$ equals ___

In triangle $ABC$: $AB = 15$, $BC = 20$, $AC = 25$.

Check: $15^2 + 20^2 = 225 + 400 = 625 = 25^2$.

So triangle $ABC$ is right-angled at $B$, meaning $\angle ABC = 90^\circ$.

Since $ABCD$ is cyclic and $\angle ABC = 90^\circ$, $AC$ is a diameter of the circle.

Therefore $\angle ADC = 90^\circ$ as well (angle in a semicircle).

In right triangle $ADC$:

$AD^2 + CD^2 = AC^2$

$AD^2 = 25^2 - 24^2 = 625 - 576 = 49$

$AD = 7$.