Cor. 1.—Since the perpendicular from O on AB bisects it [III. iii.], we see that the perpendiculars at the middle points of the sides of a triangle are concurrent.
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.