Fig. 3.

B. 1.

Cor. 2. Triangles ABC, DEF, and parallelograms AB, DE, which have equal bases AC and DF, are to one another as their altitudes.

If in the triangles one angle be right, it is evident. If not, draw BK and EG parallel to AC and DF, and through the points C and F draw CK and FG perpendicular to AC and DF, and join AK, DG.

Because the triangles ABC, AKC are upon the same base, and between the same parallels, ABC is equal to (1) Prop.87. AKC (1), and likewise DGF is equal to DEF; but if the right lines CK and FG be considered as the bases of the triangles AKC and DGF, their altitudes AC and (2) Hypoth. DF are equal (2), and therefore the triangles are as their bases CK and FG (3), therefore the triangles (3) Prec. Cor. ABC and DEF, which are equal to them, are as CK (4) Prop.34. and FG, but these lines are equal to their altitudes (4): B. 1.& Def and parallelograms are double of the triangles, and therefore are also to one another as CK and FG.

5. B. 6.

Fig. 4.

PROP. II. THEOR.

If a right line (DE) be drawn parrallel to any side (AC) of the triangle (ABC), it divides the other sides, or those sides produced, into proportional segments; and the homologous segments are at the same side of the parallel line (DE).

And if a right line (DE) divide the sides of a triangle, or those sides produced, into proportional segments, so that the homologous segments be at the same side of it, it is parallel to the remaining side (AC).

Part 1. Let DE be parallel to AC, and AD is to DB as CE is to EB.

For draw AE and DC, and since the triangles EAD (1) Prop.37. and ECD are upon the same base ED and between B. 1. the same parallels ED and CA, they are equal (1); (2) Prop.14. therefore AED has the same ratio to DEB which CDE (3) Prop. 1. has to the same EDB (2); but AED is to DEB as AD to DB (3), and CDE is to EDB as CE to EB (3), (4) Prop.18. therefore AD is to DB as CE is to EB (4).

B. 5.

B. 6.

B. 5.

Part 2. Let AD be to DB as CE to EB, and the right line DE is parallel to AC.

B. 5.

Let the same construction remain, and AD is to DB as the triangle AED to the triangle DEB (5), and (5) Prop. 1. as CE to EB, so is the triangle CDE to the triangle B. 6. EDB (5), but AD is to DB as CE to EB, therefore (6) Prop.18. AED is to DEB as CDE to the same EDB (6), there- (7) Prop.15. fore AED is equal to CDE (7): but they are upon the B. 5. same base DE, and at the same side of it, and there- (8) Prop.39. fore DE is parallel to AC (8).

Cor. If there be drawn several right lines IO and FL parallel to the same side BC of the triangle BAC, all the segments of the other sides are proportional.

B. L

Fig. 5.

Draw FQ parallel to AC. In the triangle BFQ, BI is to IF as QS to SF (1), but on account of the (1) Prop. 2. parallelograms QO and SL, CO is equal to QS, and B. 6. OL to SF (2), and therefore CO is to OL as BI to (2) Prop.34. IF; and in the same manner it can be demonstrated, whatever be the number of parallels.

PROP. III. THEOR.

B. 1.

A right line (AD) bisecting the angle of a triangle Fig. 6. (BAC) divides the opposite side into segments (BD, See N. DC) proportional to the conterminous sides (BA, AC).

And if a right line (AD) drawn from any angle of a triangle divide the opposite side (BC) into segments (BD, DC) proportional to the conterminous sides (BA, AC), it bisects the angle.

Part 1. Draw through C a right line CE parallel to AD until it meet the side BA produced to E.

Because the lines AD and EC are parallel, the angle (1) Prop.29. BAD is equal to the internal angle at the same side B. 1. AEC (1), therefore the angle DAC is equal to AEC (2) Hypoth Ar.1.B.1. (2), but DAC is equal to the alternate angle ACE (1), (9) Prop. 6. therefore ACE and AEC are equal, and therefore, the opposite sides AE and AC are equal (3); but since AD (4) Prop. 2. is parallel to EC, EA is to AB as CD is to DB (4), B. 6.

R.

B. 1.

B. 5.

therefore, since EA and AC are equal, AC is to AB as CD is to DB.

Part 2. Let the same construction remain, and BA is to AE as BD to DC (4); but BD is to DC as BA to (5) Hypoth. AC (5), therefore BA is to AE as BA to AC (6), and (6) Prop.18. therefore AE and AC are equal (7), and the angle (7)Prop.15. ACE is equal to AEC (8); but since AD and EC are B. 5. parallel, the angle DAC is equal to the alternate angle (8) Prop. 5. ACE (9), and the angle BAD equal to the internal (9) Prop.29. angle at the same side AEC (9); therefore, since AEC and ACE are equal, BAD and DAC are also equal, and therefore the right line AD bisects the angle BAC.

B. 1.

B. 1.

Fig. 7.

B. 1.

Cor. If the right line bisecting the vertical angle of a triangle bisect the base, the triangle is isosceles.

PROP. IV. THEOR.

In equiangular triangles (BAC and CDE), the sides about the equal angles are proportional; and the sides which are opposite to the equal angles are homologous.

Let the sides BC and CE, which are opposite to the equal angles BAC and CDE, be placed so that they may form one right line, the triangles be at the same side, and the equal angles BCA and ČED be not conterminous; since the angles ABC and BCA are less than (1) Prop.17. two right angles (1), and CED is equal to BCA, ABC and CED are less than two right angles, and therefore (2) Prop.12 the lines BA and ED must meet if produced (2); let them meet in F; because the angles BCA and CED (5) Hypoth. are equal (3), CA is parallel to EF (4), and because the (4) Prop. 28. angles ABC and DCE are equal, CD is parallel to BF (5) Prop.34. (4), therefore AFDC is a parallelogram, and the side AC equal to FD, and also AF equal to CD (5).

B. 1.

B. 1.

B. 1.

In the triangle BFE the line AC is parallel to FE, therefore BA is to AF, or to CD equal to AF, as BC (6) Prop. 2. to CE (6); and, by alternation, AB is to BC as CD to CE; and since CD is parallel to BF, BC is to CE as FD, or AC equal to FD, to DE (6), and, by alternation, BC is to CA as CE to ED; therefore, since AB

B. 6.

is to BC as DC to CE, and BC to AC as CE to ED,

ex æquali (7), AB is to AC as DC to DE; therefore (7)Prop 34. the sides about the equal angles are proportional, and B. 5. those which are opposite to the equal angles, are homologous.

Schol. It is evident that the triangles BAC and CDE are similar; and also that the sides, which are opposite to the equal angles, are proportional.

Cor. 1. If in any triangle ABC a parallel DE be Fig. 4. drawn to any side AC, the triangle cut off DBE is similar to the whole.

Because the angle B is common to both, and the angles BDE, BED are equal to the internal angles BAC, (1)Prop.29. BCA (1), the triangle DBE is equiangular to ABC, and therefore similar to it (2).

Cor. 2. If there be drawn a parallel DE to any side, AC, of the triangle ABC, the right line BO, drawn from the opposite angle divides the parallel lines into proportional parts.

B. 1.

(2) Schol.

prec.

Since DI is parallel to AO, AO is to DI as OB to (1) Schol. IB (1); also OC is to IE as OB to IB, therefore AO Prop. 4.B.6. is to DI as OC to IE (2), and by Alternation AO is to (2) Prop.18. OC as DI is to IE.

B. 5.

PROP. V. THEOR.

If two triangles (ABC, DEF) have their sides pro- Fig. 9. portional (BA to AC as ED to DF, and AC to CB as DF to FE) they are equiangular; and the equal angles are subtended by the homologous sides.

At the extremities of any side DE of either triangle DEF, let the angles EDG and DEG be constructed equal to the angles A and B at the extremities of the (1) Cgr. 2. side AB, which is homologous to ED, and in the Prop. 32. triangle DEG the remaining angle G is equal to the (2) Constr. angle C in the triangle ABC (1).

Because the triangles ABC and DEG are equiangular (2), BA is to AC as ED to DG (3), but BA is to AC as ED to DF (4), therefore ED is to DG as ED

B.I.

(3) Prop. 4.

B. 6
(4) Hypath.

B. 5.

(5) Prop. 18. to DF (5), and therefore DG and DF are equal (6); in B. 5. the same manner it can be demonstrated that EG and (6) Prop.15. EF are equal, therefore the triangle EDG is equilate(7) Schol. ral to EDF and therefore equiangular to it (7); but Prop. 8.B.1. the triangle EDG is equiangular to BAC (8), and (8) Constr. therefore BAC is equiangular to EDF, and it is evident that the equal sides subtend the equal angles.

Fig. 9. See N.

PROP. VI. THEOR.

If two triangles (ABC, DEF) have one angle in each equal (A equal to D) and the sides about the equal angles proportional (BA to AC as ED to DF); the triangles are equiangular, and have those angles equal which the equal sides subtend.

At the extremities of either of the sides about the equal angles DE in the triangle DEF, let the angles EDG and GED be constructed equal to the angles A and B at the extremities of the side AB, which is homologous to ED, and in the triangle DEG the re(1) Cor. 2. maining angle G is equal to the remaining angle C in the triangle ABC (1).

Prop. 32.
B. 1.

(2) Constr. (3) Prop. 4.

B. 6.

(4) Hypoth. (5) Prop.18. B. 5.

Since the triangle ABC is equiangular to DEG (2), BA is to AC as ED to DG (3); but BA is to AC as ED to DF (4), therefore ED is to DG as ED to DF (5), and therefore DG and DF are equal (6): the angles EDG and EDF are also equal, because each of them (6) Prop.15. is equal to the angle A (7), and the side ED is common B. 5. to both, therefore the triangle EDF is equiangular to (7) Hypoth. & constr. EDG (8), but BAC is equiangular to EDG (9), there(8)Prop. 4. fore BAC is equiangular to EDF; and it is evident that the equal sides subtend the equal angles.

B. 1.

(9) Constr.