Def.—The circle ABC is called the circumcircle, its radius the circumradius, and its centre the circumcentre of the triangle. Exercises. 1. The three perpendiculars of a triangle (ABC) are concurrent. Dem.—Describe a circle about the triangle. Let fall the perpendicular CF. Produce CF to meet the circle in G. Make FO = FG. Join AG, AO. Produce AO to meet BC in D. Then the triangles GFA, OFA have the sides GF, FA in one equal to the sides OF, FA in the other, and the contained angles equal. Hence [I. iv.] the angle GAF equal OAF; but GAF = GCB [III. xxi.]; hence OAF = OCD, and FOA = DOC; hence OFA = ODC; but OFA is right, hence ODC is right. In like manner, if BO be joined to meet AC in E, BE will be perpendicular to AC. Hence the three perpendiculars pass through O, and are concurrent. This Proposition may be proved simply as follows:— Draw parallels to the sides of the original triangle ABC through its vertices, forming a new triangle A′B′C′ described about ABC; then the three perpendiculars at the middle points of the sides of A′B′C′ are concurrent [v. Cor. 1], and these are evidently the perpendiculars from the vertices on the opposite sides of the triangle ABC (compare Ex. 16, Book I.). Def.—The point O is called the orthocentre of the triangle ABC. 2. The three rectangles OA.OP, OB.OQ, OC.OR are equal. Def.—The circle round O as centre, the square of whose radius is equal OA.OP = OB.OQ = OC.OR, is called the polar circle of the triangle ABC. Observation.—If the orthocentre of the triangle ABC be within the triangle, the rectangles OA.OP, OB.OQ, OC.OR are negative, because the lines OA.OP, &c., are measured in opposite directions, and have contrary signs; hence the polar circle is imaginary; but it is real when the point O is without the triangle—that is, when the triangle has an obtuse angle. 3. If the perpendiculars of a triangle be produced to meet the circumscribed circle, the intercepts between the orthocentre and the circle are bisected by the sides of the triangle. 4. The point of bisection (I) of the line (OP) joining the orthocentre (O) to the circumference (P) of any triangle is equally distant from the feet of the perpendiculars, from the middle points of the sides, and from the middle points of the distances of the vertices from the orthocentre. Dem.—Draw the perpendicular PH; then, since OF, PH are perpendiculars on AB, and OP is bisected in I, it is easy to see that IH = IF. Again, since OP, OG are bisected in I, F; IF = PG—that is, IF = the radius. Hence the distance of I from the foot of each perpendicular, and from the middle point of each side, is = the radius. In like manner, if OC be bisected in K, then IK = the radius. Hence we have the following theorem:—The nine points made up of the feet of the perpendiculars, the middle points of the sides, and the middle points of the lines from the vertices to the orthocentre, are concyclic.