The Kurrah's Theorem is described in Problem 6.10 in the Math Challenge II-B Geometry book.  Here is what it says:

Let $\triangle ABC$ be given. Let $D,E$ be points on $\overline{AC}$ (where $D$ is between $A$ and $E$) such that with $\angle ABC = \angle ADB = \angle BEC$.  Prove that
$$AB^2+BC^2 = AC(AD+CE).$$