angles ADE and DEB will have a common altitude; and consequently, we shall have, The triangles ADE and EDC have also a common altitude; and consequently, we shall have, ADE : EDC :: AE : EC; but, by hypothesis, AD : ᎠᏴ :: AE : EC ; hence (B. II., P. IV.), ADE : DEB :: ADE : ED C. The antecedents of this proportion being equal, the consequents will be equal; that is, the triangles DEB and EDC are equal. But these triangles have a common base altitudes are equal (P. VI., C.); that is, C, of the line BC, are equally distant hence, BC and DE are which was to be proved. PROPOSITION XVII. THEOREM. In any triangle, the straight line which bisects the angle at the vertex, divides the base into two segments proportional to the adjacent sides. Let AD bisect the vertical angle A of the triangle BAC: then will the segments BD and DC be 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. Then, because CE and DA are parallel, the angles BAD and AEC are equal (B. I., P. XX., C. 3); the angles DAC and ACE are E also equal (B. I., P. XX., C. 2). But, BAD and DAC are equal, by hypothesis; consequently, AEC and ACE are equal : hence, the triangle ACE is isosceles, AE being equal to A C. B D 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 will they be similar. poin', H, of BC; the point D at some point G, of BA; the side DF will take the position GH, and BGH will or, since BG is equal to ED, and BH to EF, BA : ED :: BC : EF. In like manner, it may be shown that and also, BC : EF :: CA: FD; CA : FD :: AB : DE; 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, equel to two angles in the other, each to each, they will be similar (B. L., P. XXV., C. 2). Triangles which have their corresponding sides proportional, are similar. In the triangles ABC and DEF, let the corresponding sides be proportional; that is, let AB : DE :: BC : EF : CA FD; then will the triangles be similar. For, on BA lay off BG equal to ED; on BC lay BA : BG :: BC : BH; hence, GH is parallel to AC (P. XVI.); and consequently, the triangles BAC and BGH are equiangular, and therefore similar hence, BC: BH :: CA: HG. But, BH is equal to EF; hence, HG is equal to FD. The triangles BHG and EFD have, therefore, their sides equal, each to each, and consequently, they are equal in all their parts. Now, it has just been shown that BHG and BCA are similar hence, EFD and BCA are also similar; which was to be proved. Scholium. In order that polygons may be similar, they must fulfill two conditions: they must be mutually equiangular, and the corresponding sides must be proportional. In the case of triangles, either of these conditions involves the other, which is not true of any other species of polygons. PROPOSITION XX. THEOREM. Triangles which have an angle in each equal, and the in cluding sides proportional, are similar. In the triangles ABC and DEF, let the angle B be equal to the angle E; and suppose that BA ED :: BC: EF; then will the triangles be similar. For, place the angle E upon its equal B ; F will fall at some point of BC, as II; D will fall at some point of BA, as G; B E D DF will take the position GII, and the triangle DEF will coincide with GBH, and consequently, will be equal to it. But, from the assumed proportion, and because BG is equal to ED, and BH to EF we have, BA : BG :: BC : BH ; hence, GII is parallel to AC; and consequently, BAC and BGH are mutually equiangular, and therefore similar. But, EDF is equal to BGH: hence it is also similar to BAC; which was to be proved. PROPOSITION XXI. THEOREM. Triangles which have their sides parallel, each to each, or perpendicular, each to each, are similar. 1o. Let the triangles ABC and DEF have the side AB parallel to DE, BC to EF, and CA to FD: then will they be similar. |