SIMILAR rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals.

In the case of triangles, this definition is redundant. For it is proved in Prop. IV. of this Book, that the sides about the equal angles of equiangular triangles are proportionals. In the case of quadrilaterals, or polygons, however, the definition is necessary. According to this definition, all equilateral triangles, squares, and regular pentagons are similar rectilineal figures.

II.

Triangles and parallelograms are said to have their sides reciprocally proportional, when the sides about two of their angles are proportionals in such a manner, that a side of the first figure is to a side of the second, as the remaining side of the second is to the remaining side of the first.

Two magnitudes of any kind may be said to be reciprocally proportional to other two of the same kind, when one of the first pair is to one of the second pair, as the remaining one of the second pair is to the remaining one of the first.

III.

A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.

A straight line is said to be cut in harmonical ratio, when the whole is to one of the extreme segments as the other extreme segment is to the middle segment. For brevity's sake, this mode of dividing a straight line is called harmonical section; and the mode of dividing a straight line explained in Euclid's definition is called medial section.

IV.

The altitude of any figure is the straight line drawn from its vertex perpendicular to the base.

Altitude is a term synonymous with perpendicular. By the vertex of a figure here, is meant that angular point of the figure which is most remote from any side of the figure assumed at the base, the degree of remoteness being measured by a perpendicular drawn to that side or that side produced, from the said vertex; in other words, the longest perpendicular drawn from any angular point to the base, or base produced, is the altitude.

PROP. I. THEOREM.

Triangles and parallelograms of the same altitude are to one another as their bases.

Let the triangles ABC and A CD, and the parallelograms EC and CF, have the same altitude, viz., the perpendicular drawn from the

point A to BD or BD produced. As the base B C is to the base CD, so is the triangle ABC to the triangle A CD; and as the base BC is to the base CD, so is the parallelogram EC to

the parallelogram CF.

Produce BD both ways to the points H and L, and take any number of straight lines BG and GH, each equal to the base BC (I. 3); and DK and K L, any number of straight lines each equal to the base CD. Join AĞ, AH, AK, and A L.

EA F

HG BC D K L

Because CB, B G and GH, are all equal, the triangles A H G, AGB and AB C, are all equal (I. 38). Therefore, whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle A B C. For the same reason, whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC. But if the base HC be equal to the base CL, the triangle A HC is also equal to the triangle ALC(I. 38); and if the base H C be greater than the base CL, the triangle AHC is likewise greater than the triangle ALC; and if less, less. Because there are four magnitudes, viz., the two bases BC and C D, and the two triangles A B C and ACD; and of the base B C, and the triangle A B C, the first and the third, any equimultiples whatever have been taken, viz., the base HC and the triangle AHC; and of the base CD, and the triangle ACD, the second and the fourth, any equimultiples whatever have been taken, viz., the base CL and the triangle A LC. And it has been shown, that, if the base H C be greater than the base CL, the triangls A H C is greater than the triangle ALC; if equal, equal; and if less, less. Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle A CD(V.Def.5). Because the parallelogram CE is double of the triangle A B C (I. 41), and the parallelogram CF double of the triangle A CD, and magnitudes have the same ratio which their equimultiples have (V. 15). Therefore, as the triangle A B C is to the triangle ACD, so is the parallelogram E C to the parallelogram C F. But it has been shown, that, as the base B C is to the base CD, so is the triangle A B C to the triangle A CD. And as the triangle ABC is to the triangle ACD, so is the parallelogram E C to the parallelogram C F. Therefore, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF (V. 11). Wherefore, triangles, &c. Q. E. D.

COROLLARY.-From this it is plain, that triangles and parallelograms that have equal altitudes, are to one another as their bases.

Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are (I. 33), because the perpendiculars are both equal and parallel to one another (I. 28). Then, if the same construction be made as in the proposition, the demonstration will be the same.

Exercise.-Triangles and parallelograms upon eçual bases are to one another as their altitudes.

PROP. II. THEOREM.

If a straight line be drawn parallel to one of the sides of a triangle, it cuts the other two sides, or these sides produced, proportionally, so that the segments between the base and the parallel, are homologous; and conversely, if the two sides, or these sides produced, be cut proportionally, so that the segments between the base and the parallel are homologous, the straight line which joins the points of section is parallel to the base of the triangle.

Let DE be drawn parallel to B C, one of the sides of the triangle ABC. The sides A B and A C or A B and A C produced, are cut proportionally; that is, BD is to

DA, as CE is to E A.

Join BE and CD.

The tri

B

A

CD

E

D

E B

angle BDE is equal to the triangle CDE (I. 37), because they are on the same base D E, and D between the same parallels DE, and B C. But ADE is another triangle; and equal magnitudes have the same ratio to the same magnitude (V. 7). Therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. But the triangle BDE is to the triangle ADE, as BD is to DA (VI. 1), because their altitude is the perpendicular drawn from the point E to A B, and they are to one another as their bases. For the same reason, the triangle CDE is to the triangle ADE, as CE to E A. Therefore, as BD is to D A, so is CE to EA (V. 11).

Next, let the sides A B and AC of the triangle A B C, or A B and A C produced, be cut proportionally in the points D and E, that is, so that BD is to DA as CE to E A. Join D E, and it is parallel to B C.

The same construction being made, because B D is to DA as CE is to E A. But B D is to DA, as the triangle B D E is to the triangle A DE (VI. 1); and CE is to EA, as the triangle CDE is to the triangle ADE. Therefore the triangle BDE is to the triangle AD E, as the triangle CDE is to the triangle ADE (V. 11). Wherefore the triangles B DE and CDE have the same ratio to the triangle ADE. Therefore the triangle BDE is equal to the triangle CDE (V. 9); and they are on the same base D E. But equal triangles on the same base and on the same side of it, are between the same parallels (I. 39). Therefore DE is parallel to B C. Wherefore, if a straight line, &c. Q. E. D.

The necessity for three diagrams in this proposition arises from the variety of position which may be given to the straight line drawn parallel to the base. This parallel may be drawn between the vertex of the triangle and the base, as in the first figure; beyond the base, as in the second figure, when the sides must be produced through the extremities of the base to meet it; or, beyond the vertex, as in the third figure, when the sides must be produced through the vertex, to meet it. In all these cases, the demonstration holds equally good, and it should be read with each figure separately, in order to render the argument clear to the mind. A more general method of enunciating this proposition is contained in the corollary.

Corollary.-The triangles which two intersecting straight lines form with two parallel straight lines, have their sides on the intersecting lines proportionals, and those in the same straight line are homologous; and conversely, the two straight lines, which, with two intersecting straight lines, form triangles, having

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 commor 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.

A

D C

E

Because the straight line A C 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 ACE B (I Ax. 1). Again, because the straight line BAE meets the parallels AD and EC, the exterior angle B AD is equal to the interior and opposite angle AEC (I. 29). But the angle A CE 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 (1.6). 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 DC, as BA is to AC (V. 7). Next, let the segment BD be to the segment.D C, as the side B A is to the side A C; 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 DC, as BA is to AC; and BD 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 A EC 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 opposits side, the triangle is isosceles.

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

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 A D which meets the base produced in D. The segment BI) is to the segment DC, as the side B A is to the side AC.

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

B

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 angle ACF (I. Ax. 1). Again, because the straight line FA E meets the parallels AD and FC, the exterior angle D A E 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 A F (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 B A is to the side A C, 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 A C; and BD is to D C, as B A to A F (VI. 2). Therefore B A is to A C, as BA to AF (V. 11), and AC is equal to AF (V. 9). Because the angle A FC 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 EAD is equal to the angle CAD (I. Ax. 1). Wherefore the angle CAE is bisected by the straight liv 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.