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. Let ABD be a circle circumscribed about the triangle ABC; O the centre of the inscribed circle. Join 40, 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=DẠC=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 Blet fall the perpendicular BD; and in the triangles ABD, BDC let circles be inscribed; their diameters are to one another as AB to BC. 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 CE: OF EG: 2OF: 2 EG, (30.) To find the locus of the vertex of a triangle, whose base and ratio of the other two sides are given. Let AB be the given base; divide it in C so that AC: CB may be in the given ratio of the sides. AB to 0; and take CO a mean pro Produce C B portional between AO and BO. With the centre C, and radius CO, describe a circle; it will be the locus required. In the arc AD take any point D; join DA, DB, DC, DO. Since OD=OC, AO OD OD: OB, ..the sides about the common angle O are proportional, and the triangles ADO, BDO are equiangular; :. AD : DB :: DO : OB :: CO : OB :: AO : CO :: AO-CO: CO-OB:: AC: CB, i. e. in the given ratio. In the same manner, if any other point be taken in the circumference of the circle, and lines drawn to it, they will be in the same given ratio, and .. the circumference is the locus required. COR. Since in any triangle, if from the vertex a line be drawn cutting the base in the ratio of the sides, it will bisect the angle, .. the angle ADC= BDC. (31.) A given straight line being divided into any three parts; to determine a point such, that lines drawn to the points of section and to the extremities of the line shall contain three equal angles. Let AB be the given line, and AC, CD, DB the B E given parts. Take CO a mean proportional between 40 and DO; and with the centre O and radius OC describe a circle. Produce CB; and make DE a mean proportional between CE and BE; and with the centre E, and radius ED, describe a circle cutting the former in F; F is the point required. For, as was proved in the last proposition, AF: FD :: AC : CD, and.. the angle AFC=CFD; and CF: FB: CD: DB, .. the angle CFD=DFB; and.. the three angles AFC, CFD, DFB are equal. (32.) If two equal lines touch two unequal circles, and from the extremities of them lines containing equal angles be drawn cutting the circles, and the points of section joined; the triangles so formed will be reciprocally proportional. Let two equal lines AB, CD touch two unequal circles EBF, GDH; and from A and C let lines AIK, AEF, CLM, CGH be drawn containing the equal angles KAF, MCH. Join IE, KF, GL, MH; then will the triangle AKF: CHM CGL AIE. Since AB is equal to CD, the rectangles EA, AF, and GC, CH are equal; .. AF : CH :: GC : AE, and for the same reason, AK CM: CL: AI, whence AFX AK: CH× CM:: CG × CL: AE × AI, ..the triangle AKF : CMH :: CGL : AIE, since AK CM in the ratio of the perpendiculars from K and M on AF and CH; and CL: AI in the ratio of the perpendiculars from L and I. (33.) If from an angle of a triangle a line be drawn to cut the opposite side, so that the rectangle contained by the sides including the angle, be equal to the rectangle contained by the segments of the side together with the square of the line so drawn; that line bisects the angle. From B one of the angles of the triangle ABC, let BD be drawn, so that the rectangle AB, BC may be equal to the rectangle AD, DC together with the square of BD; BD bisects the angle B. B E D For if not, let BE bisect it; the rectangle AB, BC is equal to the rectangle AE, EC together with the square of BE. About ABC describe a circle, and produce BD, BE to the circumference in F and G; join FG. The rectangle AD, DC is equal to the rectangle BD, DF; .. the rectangle AB, BC is equal to the rectangle BD, DF together with the square of BD, i. e. to the rectangle BF, BD. In the same way the rectangle AB, BC is equal to the rectangle BG, BE; |