 Looking at this diagram, we can see that all we need to prove is that the angle at vertex D of triangle ABD is a right angle. The proof is easy once we realize that since lines AC, DC, and BC are the radii of a circle, the triangles ACD and DCB are isosceles triangles, whence their base angles are equal. For then, if we denote the angles of ABD by the names of their vertices, this fact entails that D=A + B; and so, since A + B + D=180o, it follows that 2A + 2B=180o; therefore, A + B=90o, that is, D=90o, q. e. d.