(23.) The straight line bisecting any angle of a triangle inscribed in a given circle, cuts the circumference in a point, which is equidistant from the extremities of the side opposite to the bisected angle, and from the centre of a circle inscribed in the triangle. Let ABC be a triangle inscribed in the circle ACD. Bisect the angles BAC, ABC by the lines AD, BO, which meet in 0; Ois the centre of the circle inscribed in the triangle. Join BD, DC. The lines DB, DC, DO are equal to each other. B Because the angles DAB, DAC are equal, BD= DC; and because the angle CBD=CAD=DAB, to each of these add the angle CBO or its equal ABO; and the whole angle OBD is equal to the two ABO, OAB, i. e. to BOD (Eucl. i. 32.); and .. OD=DB. (24.) The perpendicular from the vertex on the base of an equilateral triangle is equal to the side of an equilateral triangle inscribed in a circle, whose diameter is the base. From C the vertex, let CO be drawn perpendicular to AB, the base of the equilateral triangle ABC; upon AB describe a circle ADB, and let DEF be an equilateral triangle inscribed in it; CO will be equal to a side of this triangle. E A G B Draw DG bisecting the angle at D, and .. bisecting EF at right angles, consequently passing through the centre. Join EG. The angles ACO, ADG, being each equal to half the angle of an equilateral triangle, are equal to each other, and AOC=DEG, each being a right angle, and AC=AB= DG, .. CO=DE. (25.) If an equilateral triangle be inscribed in a circle, and the adjacent arcs cut off by two of its sides be bisected; the line joining the points of bisection will be trisected by the sides. Let ABC be an equilateral triangle inseribed in a circle; bisect the arcs AB, AC in D and E; join DE; it is divided into three equal parts in the points F and G. E Since DE and BC cut off equal arcs BD, CE, they are parallel, and .. (Eucl. vi. 2.) AF= AG. Join BD, AE. The angle BFD=AFE, and DBF= AEF in the same segment, and BD=AE, since they subtend equal arcs; .. DF-FA. In the same manner it may be shewn that AG-GE.. Now the triangle AFG being similar to ABC is equilateral, .. DF, FG, GE are all equal, and DE is trisected. (26.) If any triangle be inscribed in a circle, and from the vertex a line be drawn parallel to a tangent at either extremity of the base; this line will be a fourth proportional to the base and two sides. Let ABC be a triangle inscribed in the circle ABC; and from B let BD be drawn parallel to AE a tan gent at A; then will AC: AB :: BC: BD. Produce CB to meet the tangent in E. Since the angle EAB is equal to the angle in the alternate segment ACB, and the A E angle AEB is equal to CBD, .. the triangle ABE is similar to CBD, and AE AB :: CB: CD; but from similar triangles BDC, EAC, (27.) If a triangle be inscribed in a circle, and from its vertex lines be drawn parallel to tangents at the extremities of its base, they will cut off similar triangles. Let ABC be a triangle inscribed in a circle, and AD, CE tangents at the points A and C. From B draw BF, BG respectively parallel to them; these lines will cut off the triangles ABF, CBG, which are similar. D B F G For (Eucl. iii. 32.) the angle ACB is equal to DAB, i. e. to the alternate angle ABF; and the angle BAC is equal to BCE, i. e. to CBG; whence the triangles ABF, CBG having two angles in each equal, will be equiangular and similar. COR. 1. The rectangle contained by the segments of the base adjacent to the angles will be equal to the square of either line drawn from the vertex. For if AD and CE be produced, they will meet and form with AC an isosceles triangle, to which BFG is similar, .. BF=BG. Now AF: BF:: BG: GC ..the rectangle AF, GC is equal to the rectangle BF, BG, i. e. to the square of BF. COR. 2. Those segments are also in the duplicate ratio of the adjacent sides. For the triangles ABF and CBG are each of them similar to ABC, whence AC: AB :: AB: AF .. AC: AF in the duplicate ratio of AC: AB; for the same reason, AC: CG in the duplicate ratio of AC: CB, :. AF : CG in the duplicate ratio of AB: CB. (28.) If one circle be circumscribed and another inscribed in a given triangle, and a line be drawn from the vertical angle to the centre of the inner, and produced to the circumference of the outer circle; the whole line thus produced has to the part produced the same ratio that the sum of the sides of the triangle has to the base. E Let ABD be a circle circumscribed about the triangle ABC; O the centre of the inscribed circle. Join AO, and produce it to D; then AOD bisects the angle BAC. Join BD, DC; and draw BO, CO to the centre of the inscribed circle; then AD: DO :: AB+AC : BC. Draw OF, OG parallel to AB, AC, meeting BD, CD in F and G. The angle DBC=DAC=DAB= DOF, and the angle at D is common to the triangles BED, OFD, and (vii. 20.) BD=DO, :. OF=be. In the same manner it may be shewn that OG=EC. Now the trapeziums BACD, FOGD being similar, and similarly situated, AD OD :: AB+ AC : FO+OG :: AB+ AC: BC. (29.) If in a right-angled triangle, a perpendicular be drawn from the right angle to the hypothenuse, and circles inscribed within the triangles on each side of it; their diameters will be to each other as the subtending sides of the right-angled triangle. Let ABC be a right-angled triangle; from the right angle B let fall the perpendicular BD; and in the triangles ABD, BDC let circles be inscribed; their diameters are to another as AB to BC. one B G E D Bisect the angles BAD, ABD by the lines AO, BO, they will meet in the centre 0; in the same manner lines bisecting DBC, DCB meet in the centre E; draw OF, EG to the points of contact. Now the triangles ABD, BDC being similar (Eucl. vi. 8.), .. the triangles ABO, BCE are similar; whence AB BC: BO: CE; but the triangles OBF, EGC are similar, .. BO .. AB CE :: OF: EG :: 2 OF 2 EG, |