197. PROP. IX. The rectangle of the segments of any chord of the circumcircle drawn through the centre of the incircle = twice the rectangle of their radii. For draw the diameter UU' of the circumcircle perpendicular to BC: and the radius IY of the incircle perpendicular to AC. Join CU'. Then, by Prop. VIII., U, I, A are in a straight line, and UI=UC. Also, UCU' in a semicircle = the right-angle IYA, and UU'C, UAC in the same segment are equal. .. the triangles UCU', IYA are similar. .. UC : UU', i.e. UI : UU' = IY : IA. .. rectangle UI, IA = rectangle UU', IY. i.e. the rectangle of the segments UI, IA = twice the rectangle of the radii of the two circles. .. the rectangle of the segments of any chord of the circumcircle drawn through I is equal to twice the rectangle of the radië. The Nine-Points Circle. 198. PROP. X. The middle points of the sides, the feet of the perpendiculars from the opposite angles, and the points midway between the angles and the orthocentre lie on a circle. Let D, E, F be the middle points of the sides of ABC; AL, BM, CN the perpendiculars from A, B, C cutting in 0; P, Q, R the middle points of A0, BO, CO. The nine points D, E, F; L, M, N; P, Q, R will lie on a circle. Draw the lines AL, CN; PE, DE. : DE bisects BC and AC, .. DE is parallel to AB. And :: PE bisects AO and AC, .. PE is parallel to OC. But AB and OC are at right-angles to one another; .. also DE and PE are at right-angles to one another. :. the circle on DP as diameter passes through E. For the same reason this circle passes through F. And it passes through L, because DLP is a right-angle. :. the circle circumscribing DEF passes through P and L. Similarly it passes through Q, M; R, and N. Hence the nine-points DEF, PQR, LMN lie on a circle. 199. PROP. XI. The gravity centre trisects and the ninepoints centre bisects the line joining the circumcentre and orthocentre. Let T the middle point of DP be the nine-points centre. (1) Produce OT to S. Let TS=TO. S shall be the circumcentre. Join SD. Then the triangles STD, OTP having two sides and an included angle at T equal, are equal in all respects. .. SD is parallel to OP and .. at right-angles to BC. A(= 2. PO=2.SD, .. AG = 2.GD. Cor. II. The angle DPL, which is the angle in the segment of the nine-points circle opposite DL, is equal to SAL; and is therefore, by Prop. VII., double of the angle between the bisector of A and the perpendicular to BC. 200. PROP. XII. The nine-points circle touches the inscribed circle. Let D bisect BC, AIH bisect A, and let IX and AL be perpendiculars upon BC. Draw HX' to touch the incircle at X'. Through D draw DX'W to cut the incircle in W. Join WL. Then (by Prop. VIII., Cor.), DH. DL= DX?. :. the angle D WL = DHX'= XIX' = 2.HIX = (by Prop. XI., Cor. II.) angle in the segment of nine-points circle opposite DL. .. (1) W is on the nine-points circle. Also the tangent at W to the nine-points circle makes with WX' an angle equal to DLW = DX'H = angle in the segment of the incircle opposite X'W. .. (2) the tangents at W to the incircle and to the nine-points circle coincide. :. the incircle and the nine-points circle touch at W. 201. In the same way it may be shown that the nine-points circle touches each of the escribed circles. For since BX,=CX, and BD = CD, .. the remainder DX, = DX. .. DX/?= DH. DL. Hence if the tangent through H is drawn to the escribed circle touching it at Xi', and a chord is drawn through DX', the other end of the chord will be the point of contact with the ninepoints circle. Now H being the centre of similitude of the inscribed and escribed circles, the tangent through H to either circle is a common tangent. Hence we see that If the fourth common tangent to the inscribed and an escribed circle be drawn, and through the points of contact chords of these circles be drawn to pass through the middle point of the side, the other extremities of the chords will be the points of contact with the nine-points circle. The Lemoine Circle. a 202. Prop. XIII. The parallels to the sides drawn through the centre of the cosine circle cut the sides in six points which lie on a circle. Construction. Let V be the middle point of KS; and the middle point of AK. Through a draw A A, antiparallel to BC. Join Va, VA Then :: V and a are the middle points of SK and AK, .. Va is parallel to SA and is half SA. But SA is perpendicular to the antiparallels to BC and is equal to the circumradius. :: Va is perpendicular to A 4, and equal to half the circumradius. Again, drawing the antiparallel to BC from K to AB, we have a line parallel to Aya and equal to the cosine-radius. |