equimultiples have (V. 15); as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF. And because it has been shewn that as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF (V. 11). COR. 1.-From this it is plain, that triangles and parallelograms that have equal altitudes are one to another as their bases. For, if the figures be placed so as to have their bases in the same straight line, then the straight line which joins the vertices is parallel to that in which their bases are (I. 34, Cor. 2). Then, if the same construction be made as in the proposition, the demonstration will be the same. COR. 2.-Rectangles, and hence also parallelograms and triangles, having equal bases, are to one another as their altitudes. If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or the other sides produced proportionally; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle. Given DE a line drawn parallel to BC, one of the sides of the triangle ABC; to prove that BD is to DA, as CE to EA. (Const.) Join BE and CD; (Dem.) then the triangle BDE because they are on the 37), is equal to the triangle CDE (I. same base DE, and between the same parallels DE and BC. But ADE is another triangle, and equal magnitudes have to the same the same ratio (V. 7); therefore, as the triangle BDE to the triangle ADE, so is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, so is BD to DA (VI. 1); because, having the same altitude, namely, the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as There the triangle CDE to the triangle ADE, so is CE to EA. fore as BD to DA, so is CE to EA (V. 11). K Next, let it be given that the sides AB and AC, of the triangle ABC, or these sides produced, are cut proportionally in the points D and E; that is, so that BD is to DA as CE to EA, and join DE; to prove that DE is parallel to BC. The same construction being made, because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE (VI. 1); and as CE to EA so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE (V. 11); that is, the triangles BDE and CDE and therefore have the same ratio to the triangle ADE; the triangle BDE is equal to the triangle CDE (V. 9). And they are on the same base DE; but equal triangles on the same base are between the same parallels (I. 39); DE is parallel to BC. therefore If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another; and if the segments of the base have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section bisects the vertical angle. Let it be given that the angle BAC of the triangle ABC is divided into two equal angles by the straight line AD; to prove that BD is to DC as BA to AC. (Dem.) Because the the angle ACE (Const.) Through the point C draw CE parallel to DA (I. 31), and let BA produced meet CE in E. straight line AC meets the parallels AD, EC, is equal to the alternate angle CAD (I. 29). But CAD is given equal to the angle BAD; wherefore BAD is equal to the angle ACE. Again, because the straight line BAE meets the parallels AD, EC, the exterior angle BAD is equal to the interior and opposite angle AEC. But the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal to the angle AEC, and consequently the side AE is equal to the side AC (I. 6). And because AD is drawn parallel to EC one of the sides of the triangle BCE, BD is to DC, as BA to AE (VI. 2). But AE is equal to AC; therefore as BD to DC, so is BA to AC (V. 7). Next, let it be given that BD is to DC as BA to AC, and join AD; to prove that the angle BAC is divided into two equal angles by the straight line AD. The same construction being made; (Dem.) because BD is to DC as BA to AC; and BD is to DC as BA to AE (VI. 2), because AD is parallel to EC; therefore AB is to AC as BA to AE (V. 11); consequently AC is equal to AE (V. 9), and the angle AEC is therefore equal to the angle ACE (I. 5). But the angle AEC is equal to the external and opposite angle BAD; and the angle ACE is equal to the alternate angle CAD. Wherefore, also, the angle BAD is equal to the angle CAD; therefore the angle BAC is cut into two equal angles by the straight line AD. If the exterior angle of a triangle be bisected by a straight line which also cuts the base produced, the segments between the bisecting line and the extremities of the base have the same ratio which the other sides of the triangle have to one another; and if the segments of the base produced have the same ratio which the other sides of the triangle have, the straight line drawn from the vertex to the point of section bisects the exterior angle of the triangle. Let it be given that the exterior angle CAE of any triangle ABC is divided into two equal angles by the straight line AD which meets the base produced in D; to prove that BD is to DC as BA to AC. E (Const.) Through C draw CF parallel to AD (I. 31). (Dem.) And because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD (I. 29); but CAD is equal to the angle DAE (Hyp.); therefore also DAE is equal to the angle ACF. Again, because the straight line FAE meets the parallels AD, FC, the exterior angle DAE is equal to the interior and opposite angle CFA. But the angle ACF has been proved to be equal to the angle DAE; therefore also the angle ACF is equal to the angle CFA; and consequently the side AF is equal to the side AC (I. 6); and because AD is parallel to FC, a side of the triangle BCF; BD is to DC as BA to AF (VI. 2); equal to AC; therefore BD is to DC as BA to ÁC. but AF is and Next, let it be given that BD is to DC as BA to AC, join AD; to prove that the angle CAD is equal to the angle DAE. The same construction being made, (Dem.) because BD is to DC as BA to AC; and also BD to DC as BA to AF; therefore BA is to AC as BA to AF (V. 11); wherefore AC is equal to AF (V. 9), and the angle AFC equal to the angle ACF (I. 5). But the angle AFC is equal to the exterior angle EAD, and the angle ACF to the alternate angle CAD; therefore also EAD is equal to the angle CAD. COR.-The two lines that bisect the vertical angle and its adjacent exterior angle cut the base produced harmonically 天 or the base is a harmonic mean between its greater internal and external segments. Schol. 1.-This and the last propositions may be enunciated thus: If the vertical angle of a triangle and its adjacent angle be bisected by lines cutting the base, it will be cut internally and externally in the ratio of the two sides. Schol. 2.-By means of these two propositions, it is proved in optics that the axis of a pencil of rays incident on a spherical mirror is divided harmonically by the radiant point, the geometrical focus of reflected rays, and the centre and surface of the reflector. It is also found that the lengths of three musical strings of the same thickness, material, and texture, and under the same tension, that produce any note, its fifth, and octave, are in harmonical progression; hence the origin of the term. It is believed that Pythagoras first observed this relation of musical strings. The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides; that is, are the antecedents or consequents of the ratios. Given ABC, DCE, two equiangular triangles, angle ABC equal to the angle DCE, having the and the angle ACB to: the angle DEC, and consequently the angle BAC equal to the sides about the equal and the angle CDE (I. 32); to prove that angles of the triangles ABC, DCE are proportionals; those sides are homologous which are opposite to the equal angles. (Const.) Let the triangle DCE be placed so that its side CE may be contiguous to BC, and in the same straight line with it;3 and because the angles ABC, ACB are together less than two right angles (I. 17), A ABC and DEC, which is equal to ACB, are also less than two right angles; wherefore BA and ED, being produced, shall meet (I. 29, Cor.); let them be produced and meet in the point F. (Dem.) And because the angle ABC is equal to the angle DCE, BF is parallel to CD (I. 28). Again, because the angle ACB is equal to the angle DEC, AC is parallel to FE; therefore FACD is a parallelogram; B E and consequently AF is equal to CD, and AC to FD (I. 34); and because AC is parallel to FF, one of the sides of the triangle FBE, BA:AF:: BC: CE (VI. 2); but AF is equal to CD; therefore (V. 7), BA: CD :: BC: CE; and alternately, AB: BC :: DC: CE. BC: CE:: FD: DE; therefore, BC:CE:: AC: DE; is parallel to BF, to AC; BC: CA: CE: ED. Again, because CD but FD is equal and alternately, Therefore, because it has been proved that AB: BC:: DC: CE; and BC: CA :: CE: ED, by equality (V. 22), BA : AC :: CD: DE. If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular, and have their angles equal which are opposite to the homologous sides. Given the triangles ABC and DEF having their sides proportionals, so that AB is to BC as DE to EF; and BC to CA as EF to FD; and consequently, by equality, BA to AC as ED to DF; to prove that the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous sides; namely, the angle ABC being equal to the angle DEF, and BCA to EFD, and also BAC to EDF. (Const.) At the points E and F, in the straight line EF, make (I. 23) the angle FEG equal to the angle ABC, and the angle EFG equal to BCA; (Dem.) wherefore the remaining angle BAC is equal to the remaining angle EGF (I. 32), and the triangle ABC is therefore equiangular to the triangle GEF; and consequently they have wherefore their sides opposite to the equal angles but there the same ratio to EF, sequently are equal (V. 9); and con for the да G same reason, DF is equal to FG; and because, in the triangles DEF, GEF, DE is equal to EG, and EF common, and also the base DF equal to the base GF; therefore (I. 8) the angle DEF is equal to the angle GEF; in the same manner it may be proved that the angle DFE is equal to GFE, and EDF to EGF; and because the. angle DEF is equal to the angle GEF, and GEF to the angle ABC; therefore the angle ABC is equal to the angle DEF; for the same reason, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D; therefore the triangle ABC is equiangular to the triangle DEF. |