188. The inscribed circle. PROP. I. The internal angular bisectors of a triangle meet in a point. B X C Let BI, CI-bisectors of B and C—cut in I. AI shall bisect A. For, draw IX, IY, IZ perpendiculars on the sides. Then, the triangles BXI, BZI having a common side BI and two angles of the one equal to two angles of the other, are equal in all respects, so that IX = IZ. Similarly IXIY, .. IY=IZ. Therefore, the triangles AZI, AYI having each a right angle and two sides of the one equal to two sides of the other, are equal in all respects, .. IA bisects A. Since IX, IY, IZ are equal and are perpendicular to the sides, I is the centre of the circle of radius IX which touches the sides of ABC internally. That is, I is the centre of the inscribed circle of ABC. 189. The Escribed Circles. PROP. II. The bisectors of two external angles and the bisector of the remaining internal angle meet in a point. Let BI1, CI-bisectors of the external angles at B and Ccut in I. AI, shall bisect A. 19 19 For, drawing perpendiculars IX1, I11, IZ1 upon the sides, we may prove precisely as in the last proposition, that IX1 = I,Y, = IZ1, and, therefore, IA bisects BAC. Since the perpendiculars from I, on the sides are equal to one another, .. I, is the centre of the circle touching AB, AC produced and BC externally. This circle is called an Escribed Circle of the triangle. COR. I. The three sides of the triangle are common tangents of the inscribed and escribed circles; and the three vertices are centres of similitude of pairs of these circles. Thus the vertex A is the external centre of similitude of the inscribed circle and the A-escribed circle, and the internal centre of similitude of the B- and C-escribed circles. 1 COR. II. If the internal bisector AII1 of A cut BC in H, since HIX, HIX, are similar triangles, IH IH= IX: IX1, .. H is the internal centre of similitude of the inscribed circle and the A-escribed circle. [Similarly, if the external bisector IAI, of A cuts BC in H1, H1 is the external centre of similitude of the B- and C-escribed circles.] COR. III. The fourth common tangent of the inscribed and the A-escribed circles passes through H and is antiparallel to BC (with respect to A). [Similarly the fourth common tangent of the B- and C-escribed circles passes through H, and is antiparallel to BC.] COR. IV. The fourth common tangent of the inscribed and the A-escribed circles cuts off from AB and AC parts, measured from A, equal to AC and AB respectively. [Similarly the fourth common tangent of the B- and C-escribed circles cuts off from BA and CA produced parts, measured from A, equal to AC and AB respectively.] COR. V. Since BI, BI, bisect two adjacent supplementary angles at B, IBI, is a right-angle. Hence the circle on II, as diameter passes through B and C, and . the perpendiculars IX, IX, from the extremities of this diameter cut off equal segments from the chord BC; 190. The Circumscribed Circle. PROP. III. The straight lines which bisect the sides of a triangle at right-angles meet in a point. Let D, E, F be the middle points of the sides of ABC. For, join AS, BS, CS. Then, the triangles BFS, AFS having FS common, and BF-AF, and the angle BFS = the angle AFS, are equal in all respects; so that BS = AS. .. the triangles BDS, CDS having the three sides of one equal to the three sides of the other, are equal in all respects. .. DS is at right-angles to BC. Since AS, BS, CS are equal to one another, .. S is the centre of the circle circumscribing ABC. If the triangle is obtuse-angled, the circumcentre falls outside the triangle, on the side remote from the obtuse angle. COR. Since the tangent at A to the circumcircle makes with AB an angle equal to the angle BCA in the opposite segment, .. it is antiparallel to BC. But the radius SA is perpendicular to the tangent at A. .. the lines SA, SB, SC are perpendicular respectively to the antiparallels to the sides. J. T. 9 191. The Orthocentre. PROP. IV. The perpendiculars to the sides from the opposite angles meet in a point. A M L Let BM, CN-perpendiculars on AC, AB—cut in O; and let AO cut BC in L. OL shall be perpendicular to BC. For, join MN. Then, the angles ANO, AMO being rightangles, a circle goes round AMON; :. NMO = NAO. And the angles BNC, BMC being right-angles, a circle goes round BNMC; :. ▲ NMB = NCB; :. ▲ NAO = NCB. .. the triangles BAL, BCN having two angles of one equal to two angles of the other, the third angles BLA, BNC are equal. But BNC is a right-angle, .. also BLA is a right-angle. The point O is called the Orthocentre of the triangle ABC. If the triangle is obtuse-angled, the orthocentre falls outside the triangle, and within the space formed by producing through the obtuse angle its two containing sides. COR. I. Since a circle goes round BNMC, .. the sides of the triangle NML are antiparallels to the sides of the triangle ABC. COR. II. Since angle NMO NAO complement of B, = = and angle LMO LCO complement of B, = .. O is the centre of the circle inscribed in LMN. COR. III. Since AB, AC are perpendicular to ON, OM respectively, .. A, B, C are the escribed centres of LMN. |