2. If the sides BC, CA, AB of the triangle ABC be denoted by a, b, c, and half their sum by s, the distances of the vertices A, B, C of the triangle from the points of contact of the inscribed circle are respectively s − a, s − b, s − c. 3. If the external angles of the triangle ABC be bisected as in the annexed diagram, the three angular points O′, O′′, O′′′, of the triangle formed by the three bisectors will be the centres of three circles, each touching one side externally, and the other two produced. These three circles are called the escribed circles of the triangle ABC. 4. The distances of the vertices A, B, C from the points of contact of the escribed circle which touches AB externally are s − b, s − a, s. 5. The centre of the inscribed circle, the centre of each escribed circle, and two of the angular points of the triangle, are concyclic. Also any two of the escribed centres are concyclic with the corresponding two of the angular points of the triangle. 6. Of the four points O, O′, O′′, O′′′, any one is the orthocentre of the triangle formed by the remaining three. 7. The three triangles BCO′, CAO′′, ABO′′′ are equiangular. 8. The rectangle CO.CO′′′ = ab; AO.AO′ = bc; BO.BO′′ = ca. 9. Since the whole triangle ABC is made up of the three triangles AOB, BOC, COA, we see that the rectangle contained by the sum of the three sides, and the radius of the inscribed circle, is equal to twice the area of the triangle. Hence, if r denote the radius of the inscribed circle, rs = area of the triangle. 10. If r′ denote the radius of the escribed circle which touches the side a externally, it may be shown in like manner that r′(s − a) = area of the triangle. 11. rr′ = s − b.s − c. 12. Square of area = s.s − a.s − b.s − c. 13. Square of area = r.r′.r′′.r′′′. 14. If the triangle ABC be right-angled, having the angle C right, 15. Given the base of a triangle, the vertical angle, and the radius of the inscribed, or any of the escribed circles: construct it. PROP. V.—Problem. To describe a circle about a given triangle (ABC). Sol.—Bisect any two sides BC, AC in the points D, E. Erect DO, EO at right angles to BC, CA; then O, the point of intersection of the perpendiculars, is the centre of the required circle. Dem.—Join OA, OB, OC. The triangles BDO, CDO have the side BD equal CD (const.), and DO common, and the angle BDO equal to the angle CDO, because each is right. Hence [I. iv.] BO is equal to OC. In like manner AO is equal to OC. Therefore the three lines AO, BO, CO are equal, and the circle described with O as centre, and OA as radius, will pass through the points A, B, C, and be described about the triangle ABC. 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.