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Book VI.

PROP. I. THEOR.

See N.

TRIANGLES and parallelograms of the same al. titude are one to another as their bases.

Let the triangles ABC, ACD, and the parallelograms EC, CF, have the same altitude, viz. the perpendicular drawn from the point A to BD: then, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF.

Produce BD both ways to the points H, L, and take any number of straight lines BG, GH, each equal to the base BC; and DK, KL, any number of them, each equal to the base CD; and

join AG, AH, AK, AL: then, because CB, BG, GH are all equal, a 38. 1. the triangles AHG, AGB, ABC are all equal a : therefore, what

ever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC: for the same reason, whatever multiple the base

E A F LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC: and if the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALC a; and if the base H G B C D K L HC be greater than the base CL, likewise the triangle AHC is greater than the triangle ALC; and if less, less : therefore, since there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD; and of the base BC and the triangle ABC, the first and third, any equimultiples whatever have been taken, viz. the base HC and triangle AHC ; and of the base CD and triangle ACD, the second and fourth, have been taken any equimultiples whatever, viz. the base CL and triangle ALC; and that it has been shown, that, if the base HC be greater than the

base CL, the triangle AHC is greater than the triangle ALC; b 5.def.5. and if equal, equal; and if less, less : therefore b, as the base BC

is to the base CD, so is the triangle ABC to the triangle ACD.

And because the parallelogram CE is double of the triangle ABCs, and the parallelogram CF double of the triangle ACD, Book VI. and that magnitudes have the same ratio which their equimultiples haved; as the triangle ABC is to the triangle ACD, so is c 41. 1. the parallelogram EC to the parallelogram CF: and because it d 15. 5. has been shown, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so ise e 11. 5. the parallelogram EC to the parallelogram CF. Wherefore, triangles, &c. Q. E. D.

Cor. From this it is plain, that triangles and parallelograms that have equal altitudes, are one to 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 f, because the f 33. 1. perpendiculars are both equal and parallel to one another: then, if the same construction be made as in the proposition, the demonstration will be the same.

PROP. II. THEOR.

IF a straight line be drawn parallel to one of the See N. sides of a triangle, it shall cut the other sides, or those 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.

Let DE be drawn parallel to BC, one of the sides of the triangle ABC: BD is to DA, as CE to EA.

Join BE, CD; then the triangle BDE is equal to the triangle CDEa, because they are on the same base DE, and between the a 37. 1. same parallels DE, BC: ADE is another triangle, and equal magnitudes have to the same the same ratiob; therefore, as the b 7.5. 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 c BD to DA, because having the same altitude, viz. c 1. 6. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the triangle

Book VI. CDE to the triangle ADE, so i sE to EA. Therefore, as BD

to DA, so is CE to EA, d 11. 5. Next, Let the sides AB, AC of the triangle ABC, or these A A

D

E

A

C

D

B

e 1. 6.

B
С D
E B

с produced, be cut proportionally in the points D, E, that is, so that BD be to DA, as CE to EA, and join DE; 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 e; and as CE to EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADĚ, as the triangle CDe to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle ADE; and therefore f the triangle BDE is equal to the triangle CDE: and they are on the same base DE; but equal triangles on the same base are between the same parallels 8; therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D.

f9. 5.

8 39.1.

PROP. III. THEOR.

See N.

IF the angle of a triangle be divided into two equal angles, 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 ano. ther, the straight line drawn from the vertex to the point of section divides the vertical angle into two equal angles.

Let the angle BAC of any triangle ABC be divided into two equal angles by the straight line AD: BD is to DC, as BA to AC. Through the point C draw CE parallel a to DA, and let BA Book VI. produced meet CE in E. Because the straight line AC meets no the parallels AD, EC, the angle ACE is equal to the alternate a 31. 1. angle CAD b: but CAD, by the hypothesis, is equal to the angle b 29. 1: BAD; wherefore BAD is equal to the angle ACE. Again, because the straight line BAE

E meets the parallels AD, EC, the outward angle BAD is equal to

A the inward 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

B

D C is equal to the side c AC; and

c 6. 1. because AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC as BA to AEd; but AE is equal d 2. 6. to AC; therefore, as BD to DC, so is BA to AC e,

e 7.5. Let now BD be to DC as BA to AC, and join AD; the angle BAC is divided into two equal angles by the straight line AD.

The same construction being made ; because, as BD to DC, so is BA to AC; and as BD to DC, so is BA to AE 4, because AD is parallel to EC ; therefore BA is to AC as BA to AEf: conse-f 11.5. quently AC is equal to AE 8, and the angle AEC is therefore g 9. 5. equal to the angle ACE b: but the angle AĚC is equal to the out-h 5.1. ward and opposite angle BAD: and the angle ACE is equal to the alternate angle CAD b: wherefore also the angle BAD is equal to the angle CAD: therefore the angle BAČ is cut into two equal angles by the straight line AD. Therefore, if the angle, &c. Q. E. D.

Book VI.

PROP. A. THEOR.

IF the outward angle of a triangle, made by produ. cing one of its sides, be divided into two equal angles, by a straight line which also cuts the base produced; the segments between the dividing 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 divides the outward angle of the triangle into two equal angles.

Let the outward angle CAE of any triangle ABC be divided into two equal angles by the straight line AD which meets the

base produced in D: BD is to DC as BA to AC. a 31.1. Through C draw CF parallel to AD a: and because the straight

line AC meets the parallels AD, FC, the angle ACF is equal to b 29. 1. the alternate angle CADb: but CAD is equal to the angle c Hyp. DAE <; therefore also DAE is equal to the angle ACF. Again,

because the straight line FAE meets the parallels AD, FC, the outward angle DAE is equal

E
to the inward and opposite
angle CFA: but the angle

A
ACF has been proved equal
to the angle DAE; therefore
also the angle ACF is equal
to the angle CFA, and conse-
quently the side AF is equal B

С

D d 6. 1. to the side ACd: and because AD is parallel to FC, a side of the triangle BCF, BD

is to DC as e 2. 6. BA to Afe; but AF is equal to AC; as therefore BD is to DC,

so is BA to AC.

Let now BD be to DC as BA to AC, and join AD; the angle CAD is equal to the angle DAE.

The same construction being made, because BD is to DC f11. 5. as BA to AC; and that BD is also to DC as BA to Aff; 8 9. 5.

therefore BA is to AC as BA to AF 8; wherefore AC is equal h 5. 1. to AF h, and the angle AFC equal li to the angle ACF : but

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