Let ABC, A′B′C′ (Fig. 4) be two triangles equiangular in the same sense. Let BC, B′C′ meet in X. Describe circles round BXB′, CXC′ to meet again in O. Then it is easy to see that the triangles BOC, COA, AOB are equiangular in the same sense to the triangles B′OC′, C′OA′, A′OB′ respectively. Hence the triangles AOA′, BOB′, COC′ are similar;

∴ a. AA′, b. BB′, c. CC′ are proportional to a. AO, b. BO, c. CO, where a, b, c are the sides of the triangle ABC.