That this is true is in our case easily seen. We may suppose that the triangles have a common vertex, and their bases in the same line. We set off the base a along the line containing the bases m times; we then join the different parts of division to the vertex, and get m triangles all equal to [alpha]. The triangle on ma as base equals, therefore, m[alpha]. If we proceed in the same manner with the base b, setting it off n times, we find that the area of the triangle on the base nb equals nß, the vertex of all triangles being the same. But if two triangles have the same altitude, then their areas are equal if the bases are equal; hence m[alpha] = nß if ma = nb, and if their bases are unequal, then that has the greater area which is on the greater base; in other words, m[alpha] is greater than, equal to, or less than nß, according as ma is greater than, equal to, or less than nb, which was to be proved. Entry: 49