COROLLARY 3. THEOREM. If neither the bases nor altitudes of triangles and parallelograms are equal, they are to one another in the compound ratio of their bases and altitudes. PROPOSITION II. THEOREM [1.]-If a straight line (DE) be parallel to the base (BC) of a triangle (ABC), it cuts the other sides, or those sides produced, so that their segments between the parallel and the base (BD and CE) have the same ratio to their segments between the parallel and the vertex (DA, EA); that is, BD is to DA, as CE is to EA. [2.]-In a triangle (ABC) if the sides, or sides produced, be cut by a straight line (DE), so that their segments between the straight line and the base (BD, CE) have the same ratio as their segments between the straight line and the vertex (DA, EA); the straight line is parallel to the base. CONSTRUCTION. Join BE, CD. DEMONSTRATION. [1.] The triangles BDE and CDE are equal, because they are on the same base DE, and between the same E parallels DE and BC (a). Now, ADE is another triangle, and equal magnitudes have to the same the same ratio (b), therefore the triangle BDE is to the triangle ADE, as the triangle CDE is to the triangle ADE: but as the triangle BDE is to the triangle ADE, so is BD to DA, because, having the same altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases (c); and for the same reason, as the triangle CDE is to the triangle ADE, so is CE to EA: therefore, as BD is to DA, so is CE to EA (d). [2.] Because BD is to DA, as CE is to EA (e); and BD is to DA, as the triangle BDE is to the triangle ADE (c); and CE is to EA, as the triangle CDE is to the triangle ADE (c): therefore, the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE (d); that is, the triangles BDE, CDE have the same ratio to the triangle ADE; therefore the triangle BDE is equal to the triangle CDE (ƒ); and they are on the same base : but equal triangles on the same base are between the same parallels (g), therefore DE is parallel to BC. SCHOLIUM. This proposition consists of two distinct theorems, each the converse of the other. The enunciation of this proposition, as given by Simson, is very defective, inasmuch as he omits to state which of the segments of the sides are homologous to one another in the proportion: and that of the converse theorem is, strictly speaking, false, since a straight line may cut the sides of a triangle proportionally, without being parallel to the base; as in the figure, where AD is to DB, as CE is to EA. The necessity for three figures in the foregoing proposition arises from the varying position which the line DE may have in reference to the triangle, viz. beyond either the vertex or the base, or between them. COROLLARY. THEOREM. If several parallels (DE, FG, HI) be drawn to the base of a triangle (ABC), every pair of corresponding segments in each side will be proportional; that is, as AD is to DF, so is AE to EG, and as FH is to HB, so is GI to IC. DEMONSTRATION. For, draw HK and FL parallel to AC. Then in the parallelograms FI, MC, the opposite sides are equal (a), therefore, FM equal GI, and ML equal IC; and in the triangles AFG and FBL, AD is to DF, as AE is to EG (b); and FH is to HB, as FM is to ML (b), that is, as GI is to IĆ. H F M PROPOSITION III. THEOREM [1.]-If the angle of a triangle (ABC) be bisected by a straight line (AD) which also cuts the base, the segments of the base (BD, DC) shall have the same ratio which the other sides of the triangle (AB, AC) have to one another. [2.] And if a straight line (AD) drawn from any angle of a triangle (ABC) divide the opposite side into segments (BD, DC) which have the same ratio as the adjacent sides (AB, AC), it bisects the angle. CONSTRUCTION. Through C draw CE parallel to DA (a); then BA produced will meet CE (b), let them meet in E. : DEMONSTRATION. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD (c) but CAD, by the hypothesis, is equal to the angle BAD: therefore the angle BAD is equal to the angle_ACE (d). Again, because the straight line BAE meets the parallels AD, EC, the external angle BAD is equal to the internal and opposite angle AEC (c): but the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal to the angle AEC (d), D (a) I. 31. to I. 29. E 2 and consequently the side AE is equal to the side AC (e): and because AD is drawn parallel to EC, one of the sides of the triangle BCE; therefore BD is to DC, as AB is to AE (f): but AE is equal to AC; therefore BD is to DC, as AB is to AC. [2.] Because AD is parallel to EC, BD is to DC, as AB is to AE (ƒ); and BD is to DC, as AB is to AC (g); therefore AB is to AC, as AB is to AE (h); consequently AC is equal to AE (i), and therefore the angle AEC is equal to the angle ACE (k); but the angle AEC is equal to the external and opposite angle BAD; and the angle ACE is equal to the alternate angle CAD (c); wherefore also the angle BAD is equal to the angle CAD (d); that is, the angle BAC is bisected by the straight line AD. COROLLARY. From this proposition it follows that if the same straight line bisects the angle and the base the triangle is isosceles. PROPOSITION A. THEOREM [1.]-If an exterior angle of a triangle (ABC) be bisected by a straight line (AD) which also cuts the base produced, the segments between the bisecting line and the extremities of the base (DB, DC), have the same ratio to one another, as the other sides of the triangle (AB, AC) have. [2.] And if the segments (BD, DC) of the base produced, have the same ratio which the other sides of the triangle (AB, AC) have, the straight line (AD) drawn from the vertex to the point of section bisects the exterior angle (CAE) of the triangle. DEMONSTRATION [1.] Let the exterior angle CAE of any triangle ABC, be bisected by the straight line AD which meets the base produced in D; then BD is to DC, as BA is to AC. For, through C draw CF parallel to AĎ (a); and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD (b): but CAD is equal to the angle DAE (c); therefore also DAE is equal to the B 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 (d); and because AD is parallel to FC, a side of the triangle BCF, BD is to DC, as BA is to AF (e); but AF is equal to AC; therefore, as BD is to DC, so is BA to AC. [2.] Next let BD be to DC, as BA is to AC, and join AD; then the angle CAD is equal to the angle DAE. The same construction being made, because BD is to DC, as BA is to AC; and also BD to DC, as BA is to AF (e); therefore BA is to AC, as BA is to AF (ƒ); wherefore AC is equal to AF (g) ; and the angle AFC equal to the angle ACF (h): 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. SCHOLIA. 1. This proposition consists of two theorems, the converse of each other, and is really only a second case of the third proposition. It was inserted by Dr. Simson, who imagines it to have been omitted from the Elements by some unskilful editor. 2. When the triangle ABC is isosceles, the line from the vertex which bisects the exterior angle is parallel to the base. COROLLAPY. If both the exterior angle (CAE) and the adjacent interior angle (BAC) of a triangle be bisected by straight lines (AD and AG) which cut the base and its production, the base C E thus produced is harmonically divided, that is, BD, DG, and DC are in harmonical proportion. |