their sides on the intersecting straight lines proportionals, and those in the same straight line homologous, are parallel.

Exercise 1-If a straight line be drawn parallel to the base and cutting the sides of of a triangle, these sides are proportional to the segments cut off each of them respectively; and in the two triangle thus formed, the sides about the common angle, or the vertical angles, are proportionals.

Exercise 2.-If several straight lines be drawn parallel to the base and cutting the sides of a triangle, the segments of the sides intercepted between the same parallels are proportional to each other, and to the sides from which they are respectively cut off.

PROP. III. THEOREM.

If any angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base have the same ratio to one another which the adjacent sides of the triangle have; and conversely, if the segments of the base have the same ratio to one another which the adjacent sides of the triangle have, the straight line drawn from the vertex to the point of section bisects the vertical angle.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD. The segment BD is to the segment D C, as the side BA is to the side A C.

Through the point C draw CE parallel to DA (I. 31); and let B A produced meet CE in E.

D

E

Because the straight line AC meets the parallels AD and EC, the angle ACE is equal to the alternate angle CAD (I. 29). But the angle CAD (Hyp.) is equal to the angle BAD. Therefore the angle BAD is equal to the angle A CE (I Ax. 1). Again, because the straight line BAE meets the parallels AD and E C, the exterior angle BAD is equal to the interior and opposite angle A EC (I. 29). But the angle ACE has been proved equal to the angle BAD. Therefore also the angle ACE is equal to the angle AEC (I. Ax. 1), and the side A E to the side A C (I. 6). Because AD is drawn parallel to EC, one of the sides of the triangle BCE. Therefore BD is to DC, as BA to AE (VI. 2). But A E is equal to A C. Therefore, BD is to D C, as BA is to AC (V. 7).

Next, let the segment BD be to the segment DC, as the side BA is to the side AC; and let AD be joined. The angle B A C is bisected by the straight line A D.

The same construction being made; because BD is to D C, as BA is to AC; and B D is to DC, as BA is to AE, because A D is parallel to EC (VI. 2). Therefore BA is to A C, as BA is to AE (V. 11), and AC is equal to AE (V.9). Because the angle AEC is equal to the angle ACE (I. 5). But the angle AEC is equal to the exterior and opposite angle BAD; and the angle ACE is equal to the alternate angle CAD (I. 29). Therefore the angle BAD is equal to the angle CAD (I. Ax. 1). Wherefore the angle B A C is bisected by the straight line AD. Therefore, if the angle, &c. Q. E. D.

Corollary.-If the same straight line bisect an angle of a triangle and its opposite side, the triangle is isosceles.

Exercise.-The straight line which bisects any angle of a triangle and likewise cuts the base (that is, the interior bisecting line), divides the triangle into two triangles which are to one another as their sides which contain the bisected angle.

PROP. A. THEOREM.

If the exterior angle of a triangle be bisected by a straight line which cuts the base produced, the segments between the bisecting line and the extremities of the base, have to one another the same ratio which the adjacent sides of the triangle have; and conversely, if the segments of the base produced have the same ratio to one another which the adjacent 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 A B C be a triangle, having one of its sides B A produced to E; and let the outward angle CAE be bisected by the straight line AD which meets the base produced in D. The segment BD is to the segment DC, as the side B A is to the side A C.

Through the point C, draw CF parallel to AD (I. 31).

F

E

Because the straight line AC meets the parallels AD and FC, the angle A C F is equal to the alternate angle CAD (I. 29).__ But the angle CAD is equal to the angle DAE (Hyp.). Therefore the angle DAE is equal to the angie ACF (I. Ax. 1). Again, because the straight line FAE meets the parallels AD and FC, the exterior angle DAE is equal to the interior and opposite angle CFA (I. 29). But the angle ACF has been proved equal to the angle DAE. Therefore the angle ACF is equal to the angle CFA (I. Ax. 1), and the side AF to the side A C (I. 6). Because A D is parallel to F C, a side of the triangle BCF. Therefore BD is to DC, as BA to AF (VI. 2). But AF is equal to A C. Therefore B D is to DC, as BA is to AC (V. 7).

Next, let the segment BD be to the segment D C, as the side BA is to the side AC, and let AD be joined. The angle CAE is bisected by the straight line A D.

The same construction being made, because BD is to DC, as BA to AC; and BD is to D C, as B A to AF (VI. 2). Therefore BA is to AC, as BA to AF (V. 11), and AC is equal to AF (V. 9). Because the angle AFC is equal to the angle ACF (I. 5). But the angle AFC is equal to the exterior angle EAD (I. 29), and the angle ACF to the alternate angle CAD. Therefore E A D is equal to the angle CAD (I. Ax. 1). Wherefore the angle CAE is bisected by the straight line AD. Therefore, if the exterior, &c. Q. E. D.

This very useful proposition was added to Book VI., by Dr. Simson. It is generally considered another case of Prop. III., and it might have been incorporated with that proposition.

When the triangle ABC is isosceles, the straight line AD, which bisects the exterior angle at the vertex, is then parallel to the base, and the proposition fails. In all other cases, this exterior bisecting line cuts the base produced either on the same side with the exterior angle, or on the opposite side.

Corollary 1.-The circle described on the straight line intercepted between the points where the interior and exterior bisecting lines cut the base, passes through the vertex of the triangle.

Corollary 2.--The straight line intercepted between the point where the exterior bisecting line cuts the base produced, and the remote extremity of the base, is harmonically divided at the near extremity of the base, and the point where the interior bisecting line cuts the base.

PROP. IV. THEOREM.

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.

F

Let A B C and DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle A CB to the angle DEC; and consequently the angle B A C equal to the angle CDE (I. 32). The sides about the equal angles of the triangles A B C and DCE are proportionals; and the sides which are opposite to the equal angles are homologous. Let the triangle DCE be placed, so that its side CE may be contiguous to BC, and in the same straight line with it (I. 22).

A

B

CE

Because the angle B CA is equal to the angle CED (Hyp.). To each of these equals, add the angle ABC. Therefore the two angles A B C and BCA are equal to the two angles ABC and CED (I. Ax. 2). But the two angles A B C and BCA are together less than two right angles (1.17). Therefore the two angles ABC and CED are also less than two right angles, and BA and ED if produced will meet (I. Ax. 12). Let them be produced and meet in the point F. Because the angle ABC is equal to the angle DCE (Hyp.), BF is parallel to CD (I. 28). Because the angle ACB is equal to the angle DEC, AC is parallel to FE (I. 28). Therefore FACD is a parallelogram; and AF is equal to CD, and AC to FD (I. 34). Because AC is parallel to FE, one of the sides of the triangle F B E, BA is to AF, as B C to CE (VI. 2). But AF is equal to CD. Therefore BA is to CD, as BC is to CE (V. 7); and alternately, AB is to BC as DC is to CE (V. 16). Again, because CD is parallel to BF, BC is to CE, as FD is to DE (VI. 2). But F D is equal to A C. Therefore, BC is to CE as AC is to D E (V. 7); and alternately, B C is to CA, as CE is to ED (V. 16). Because it has been proved that AB is to BC, as DC to CE, and B C is to CA, as CE to ED. Therefore, ex æquali, BA is to AC as CD is to DE (V. 22). Therefore the sides, &c. Q. E. D.

This important proposition might be more easily demonstrated, by cutting from A B and A C parts equal to DC and DE, and joining the points where these parts are cut off. The triangle thus formed can be shown to be equal to the triangle DCE, and their angles are respectively equal. It can then be shown that the sides of this triangle are proportional to the sides of the triangle ABC about the common angle A. As the same construction can be made for each of the angles B and C, it is at once inferred that the sides about each of the angles of equiangular triangles are proportionals. To write out this construction and demonstration will be a useful exercise for the student.

Corollary 1.-Equiangular triangles are similar figures, and the sides opposite two equal angles are homologous.

Corollary 2.-The sides opposite two equal angles of equiangular triangles, are proportionals.

Corollary 3.-Triangles that are similar to the same triangle, are similar to one another.

Corollary 4.-Isosceles triangles which have one angle in the one equal to one angle in the other are similar.

Corollary 5.-A parallel to the base of a triangle cuts off from it a similar triangle.

Exercise 1.-In a triangle, the straight line drawn from the vertex bisecting the base, bisects every parallel to the base intercepted by the sides. Exercise 2.-In equiangular triangles, the perpendiculars drawn from the vertices of equal angles to the opposite sides, are proportional to those sides.

PROP. V. THEOREM.

If the sides of two triangles, about each of their angles, be proportionals, the triangles are equiangular; and the equal angles are those which are opposite to the homologous sides.

Let the triangles A B C and DEF have their sides proportionals, sɔ that A B is to B C, as D E to EF; and BC is to CA, as EF to FD; and therefore, ex æquali, B A is to A C, as ED is to DF. The triangle ABC is equiangular to the triangle DEF, and the angles which are opposite to the homologous sides are equal, viz. the angle ABC equal to the angle DEF, the angle BCA to the angle EFD, and the angle BAC to the angle EDF.

At the points E and F, in the straight line EF, make the angle FEG equal to the angle ABC, and the angle EF G equal to BCA (I. 23).

Because the remaining angle EGF, is equal to the remaining angle BAC (I. 32), the triangle GEF is equiangular to the triangle ABC. There

A

D

F

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fore they have their sides opposite to the equal angles proportionals (VI. 4). Wherefore, A B is to BC as GE is to EF. But A B is to B C.

as DE is to EF (Hyp.). Therefore DE is to EF, as GE is to EF (V. 11). Because D E and GE have the same ratio to EF, DE is equal to GE (V. 9). For the same reason, DF is equal to F G. Because, in the triangles DEF and GEF, DE is equal to E G, and EF is common, the two sides DE and EF are equal to the two GE and EF, each to each. But the base D F is equal to the base G F. Therefore the angle DEF is equal to the angle GEF (I. 8), and the remaining angles of the one to the remaining angles of the other, each to each. Therefore the angle DFE is equal to the angle GFE, and the angle EDF to the angle EGF. Because the angle DEF is equal to the angle G EF, and the angle GEF is equal to the angle ABC (Const.). Therefore the angle ABC is equal to the angle DEF (I. 4x. 1). For the same reason, the angle A C B is equal to the angle DFE, and the angle at A is equal to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q. E. D.

This proposition is the converse of Prop. IV. The 4th and 5th of Book VI., and the 47th and 48th of Book I., constitute the most important principles in the Elements. They include the principles of Trigonometry and its various applications, as well as those of analytic geometry in general.

PROP. VI. THEOREM.

If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals; the triangles are equiangular, and those angles are equal which are opposite to the homologous sides.

Let the triangles ABC and DEF have the angle BAC in the one.

K

equal to the angle EDF, in the other; and the sides about those angles proportionals, that is, BA to AC, as ED to DF. The triangles ABC and DEF are equiangular, and the angle ABC is equal to the angle DEF, and the angle ACB to the angle DFE.

At the points D and F', in the straight line DF, make the angle FDG equal to either of the angles BAC or EDF (I. 23); and the angle DFG ecual to the angle ACB.

A

D

CE

F

Because the remaining angle at B is equal to the remaining angle at G (I. 32), the triangle B DGF is equiangular to the triangle. ABC. Therefore BA is to AC, as GD is to DF (VI. 4). But (Hyp.) BA is to AC, as ED is to DF. Therefore ED is to DF, as GD is to DF (V. 11); and ED is equal to DG (V. 9). Because ED is equal to DG and DF is common to the two triangles EDF and GDF. Therefore the two sides ED and DF are equal to the two sides GD and DF, each to each. But the angle EDF is equal to the angle GDF (Const.). Therefore the base EF is equal to the base FG (I. 4), and the triangle EDF to the triangle GDF. Because the remaining angles of the one are equal to the remaining angles of the other, each to each. Therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E. But the angle DFG is equal to the angle A CB (Const.). Therefore the angle ACB is equal to the angle DFE (I. Ax. 1). But the angle BAC is equal to the angle EDF (Hyp.). Therefore the remaining angle at B is equal to the remaining angle at E (I. 32); and the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D.

Corollary 1.-Triangles which have a common angle and the sides about it proportionals, have their bases parallel, and are equiangular and similar.

Corollary 2.-Triangles which have a common angle and parallel bases, are equiangular and similar.

Exercise.-If from any number of points in a straight line parallels be drawn proportional to the distances or these points from a given point, the straight line joining this point and the extremity of one of the parallels passes through the extremities of all the parallels.

If two triangles have one angle of the cne equal to one angle of the other, and the sides about another angle in each, proportionals; and if the remaining angle in each be of the same affection (that is, either both acute, or both not acute); the two triangles are equiangular and similar.

Let the two triangles ABC and DEF have one angle BAC in the one equal to the angle EDF in the other, and the sides about their two other angles ABC and DEF, proportionals, so that A B is to BC as DE is to EF; and let their remaining angles ACB and DFE be of the same affection, that is, either both acute, or both not acute. The triangles A B C and D E F are equiangular and similar.

For if the angle ABC is not equal to the angle DEF one of them must be greater than the other. Let ABC be the greater angle, and at the point B in the straight line A B, make the angle ABG equal to the angle D E F (I. 23).