PROPOSITION XX. THEOREM. Triangles which have an angle in each equal, and the including sides proportional, are similar. In the triangles ABC and DEF, let the angle B be equal to the angle E; and suppose that ВА 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 H; D will fall B нс at some point of BA, as E D 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, ВА : BG :: BC BH; : But, hence, GH is parallel to AC; and consequently, BAC and BGH are equiangular, and therefore similar. 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. and DEF have the side 1o. Let the triangles ABC and AB parallel to DE, BC to EF, and CA to FD: then will they be similar. For, since the side AB is parallel to DE, to EF, the angle B is equal to the angle E XXIV.); in like manner, the angle C is equal to A and BC (B. I., P. D the angle F, and the an gle A to the angle D; the triangles are, therefore, mutually equiangular, and B E consequently, are similar (P. XVIII.) ; which was to be proved. 2o. Let the triangles ABC and DEF have the side AB perpendicular to DE, BC to EF, and CA to FD then will they be similar. For, prolong the sides of the triangle DEF till they meet the sides of the triangle ABC. The sum of the interior angles of the quadrilateral BIEG is equal to four right angles (B. I., P. XXVI.); but, the angles EIB and EGB are each right B H angles, by hypothesis; hence, the sum of the angles IEG IBG is equal to two right angles; the sum of the angles IEG and DEF is equal to two right angles, because they are adjacent; and since things which are equal to the same thing are equal to each other, the sum of the angles IEG and IBG is equal to the sum of the angles IEG and DEF; or, taking away the common part IEG, we have the angle IBG equal to the angle DEF. In like manner, the angle GCH may be proved equal to the angle EFD, and the angle HAI to the angle EDF; the triangles ABC and DEF are, therefore, mutually equiangular, and consequently similar; which was to be proved. Cor. 1. In the first case, the parallel sides are homolo gous; in the second case, the perpendicular sides are homologous. Cor. 2. The homologous angles are those included by sides respectively parallel or perpendicular to each other. Scholium. When two triangles have their sides perpendicular, each to each, they may have a different relative position from that shown in the figure. construct a triangle within the triangle But we can always ABC, whose sides shall be parallel to those of the other triangle, and then the demonstration will be the same as above. PROPOSITION XXII. THEOREM. ५ If a line be drawn parallel to the base of a triangle, and lines be drawn from the vertex of the triangle to points of the base, these lines will divide the base and the parallel proportionally. Let ABC be a triangle, BC its base, A its vertex, DE parallel to BC, and AF, AG, AH, lines drawn from A to points of the base: then will DI : BF :: IK : FG :: KL : GH :: LE: HC. DI : BF :: IK : FG :: KL : GH :: LE : HC; which was to be proved. Cor. If BC is divided into equal parts at F, G, and H, then will DE be divided into equal parts, at I, K, and L. PROPOSITION XXIII. THEOREM. If, in a right-angled triangle, a perpendicular be drawn from the vertex of the right angle to the hypothenuse: The triangles on each side of the perpendicular will be similar to the given triangle, and to each other: 2o. Each side about the right angle will be a mean proportional between the hypothenuse and the adjacent segment: 3°. The perpendicular will be a mean proportional between the two segments of the hypothenuse. 1o. Let ABC be a right-angled triangle, A the vertex of the right angle, BC the hypothenuse, and AD perpendicular to BC then will ADB and ADC be similar to ABC, and consequently, similar to each other. The triangles ADB and ABC have the angle B common, B D and the angles ADB and BAC equal, because both are right angles; they are, therefore, similar (P. XVIII., C). In like manner, it may be shown that the triangles ADC and ABC ABC are similar; and since ADB and ADC are both similar to ABC, they are similar to each other; which was to be proved. 2o. AB will be a mean proportional between BC and BD; and AC will be a mean proportional between CB and CD. For, the triangles ADB and B BAC being similar, their homologous sides are proportional: hence, ВС : AB :: AB : BD. In like manner, BC : AC :: AC DC; which was to be proved. A D 3o. AD will be a mean proportional between BD and DC. For, the triangles ADB and ADC being similar, their homologous sides are proportional; hence, |