In any triangle, a line drawn parallel to the base divides the other sides proportionally. Let ABC be a triangle, and DE a line parallel to the base BC: then AD : DB :: AE : EC. Draw EB and DC. Then, because the triangles AED and DEB have their bases in the same line AB, and their vertices at the same point E, they have a common altitude: hence (P. VI., C.), AED : DEB :: AD : DB. B The triangles AED and EDC, have their bases in the same line AC, and their vertices at the same point D; they have, therefore, a common altitude; hence, AED : EDC :: AE : EC. But the triangles DEB and EDC have a common base DE, and their vertices in the line BC, parallel to DE: they are, therefore, equal: hence, the two preceding proportions have a couplet in each equal; and consequently, the remaining terms are proportional (B. II., P. IV.), hence, AD : DB :: AE : EC; which was to be proved. Cor. 1. We have, by composition (B. II., P. VI.), AD + DB : AD :: AE + EC : AE; Cor. 2. If any number of parallels be drawn cutting two lines, they divide the lines propor tionally. For, let O be the point where AB and CD meet. In the triangle OEF, the line AC being parallel to the base EF, we have, OE : AE :: OF : CF. In the triangle OGH, we have, OE : EG OF : FH; :: If a straight line divides two sides of a triangle proportionally, it is parallel to the third side. Let ABC be a triangle, and let DE divide AB and AC, so that AD : DB :: AE : EC; then DE is parallel to BC. Draw DC and EB. Then the triangles D GEOMETRY. ADE and DEB have a common altitude and consequently, we have, ADE : DEB :: AD: DB. The triangles ADE and EDC have also a common altitude; and consequently, we have, ADE : EDC :: AE : EC; but, by hypothesis, AD : DB :: AE : EC; hence (B. II., P. IV.), ADE : DEB :: ADE EDC. D A The antecedents of this proportion being equal, the consequents are equal; that is, the triangles DEB and EDC are equal. But these triangles have a common base DE: hence, their altitudes are equal (P. VI., C.); that is, the points B and C, of the line BC, are equally distant from DE, or DE prolonged: hence, BC and DE are parallel (B. I., P. XXX., C.); which was to be proved. In any triangle, the straight line which bisects the angle at the vertex, divides the base into two segments propor. tional to the adjacent sides. Let AD bisect the vertical angle A of the triangle BAC: then the segments BD and DC are proportional to the adjacent sides BA and CA. From C, draw CE parallel to DA, and produce it until it meets BA prolonged, at E. AC. Then, because CE and DA and AEC are equal (B. I., B In the triangle BEC, the line AD is parallel to the base EC: hence (P. XV.), BA : AE :: BD : DC; or, substituting AC for its equal AE, BA : AC :: BD : DC; which was to be proved. PROPOSITION XVIII. THEOREM. Triangles which are mutually equiangular, are similar. Let the triangles ABC and DEF have the angle A equal to the angle D, the angle B to the angle E, and the angle C to the angle F: then they the point F fall at some point H, of BC; the point Dat some point G, of BA; the side DF will take the position GH, and BGH will be equal to EDF. Since the angle BHG is equal to BCA, GH will be parallel to AC (B. I., P. XIX., C. 2); and consequently, we have (P. XV.), H C E F BA : BG : BC : BH; or, since BG is equal to ED, and BH to EF, hence, the sides about the equal angles, taken in the same order, are proportional; and consequently, the triangles are similar (D. 1); which was to be proved. Cor. If two triangles have two angles in one, equal to two angles in the other, each to each, they are similar (B. I., P. XXV., C. 2). PROPOSITION XIX. THEOREM. Triangles which have their corresponding sides proportional, are similar. In the triangles ABC and DEF, let the corresponding sides be proportional; that is, let |