This is another generalization that does not require right angles. It's due to Thâbit ibn Qurra (836-901) [Eves]. If angles CAB, AC'B and AB'C are equal then AC² + AB² = BC(CB' + BC'). Indeed, triangles ABC, AC'B and AB'C are similar. Thus we have AB/BC' = BC/AB and AC/CB' = BC/AC which immediately leads to the required identity. In case the angle A is right, the theorem reduces to the Pythagorean proposition and proof #6.
The same diagram is exploited in a different way by E. W. Dijkstra who concentrates on comparison of BC with the sum CB' + BC'.