Book VI.

D

Let the triangle DCE be placed, fo that its fide CE may be contiguous to BC, and in the fame ftraight line with it: And because the angles ABC, ACB are together less the b 17. 1. two right angles b, ABC and DEC, which is equal to ACB, are alfo lefs than two right angles: wherefore BA, ED produced shall F cCor. 29. 1. meet c; let them be produced and meet in the point F; and because the angle ABC is equal to the angle A d 28. 1. DCE, BF is parallel d to CD. Again, because the angle ACB is equal to the angle DEC, AC is parallel to FEd: Therefore FACD is a parallelogram; and confequently AF is equal to CD, and AC to B FDe And because AC is parallel to FE, one of the fides of the triangle FBE, BA: AF:: f2.6. BC: CE f: but AF is equal to CD; therefore g, BA : CD :: 87. 5. BC: CE; and alternately, BA: BC: DC: CEb: Again, be. cause CD is parallel to BF, BC:CE::FD: DE f; but FD is equal to AC; therefore, BC: CE :: AC: DE; and alternately, BC: CA::CE: ED. Therefore, because it has been proved that AB: BC::DC: CE; and BC: CA:: CE: ED, ex æquali, BA: AC:: CD: DE. Therefore the fides, &c. Q. E. D.

e 34. 1.

C

E

PROP. V. THEOR.

F the fides of two triangles, about each of their angles, be proportionals, the triangles fhall be equiangular, and have their equal angles oppofite to the homologous fides.

Let the triangles ABC, DEF have their fides proportionals, so that AB is to BC, as DE to EF; and BC to CA, as EF to FD; and confequently, ex æquali, BA to AC, as ED to DF; the triangle ABC is equiangular to the triangle DEF, and their equal angles are oppofite to the homologous fides,

viz,

viz. the angle ABC being equal to the angle DEF, and BCA Book VI.

to EFD, and alfo BAC to EDF.

pofite to the equal angles proportionals c. Wherefore, c 4. 6. AB: BC::GE: EF; but by fuppofition,

AB: BC: DE: EF, therefore,

DE: EF::GE: EF.

Therefore d DE and GE d H. §. have the fame ratio to EF, and confequently are equal e. e 9. 5. For the same reafon, DF is equal to FG: And because, in the triangles DEF, GEF, DE is equal EG, and EF common, and alfo the bafe DF equal to the bafe GF; therefore the angle DEF is equalf to the angle GEF, and the other fs. 1, angles to the other angles, which are fubtended by the equal fides g. Wherefore the angle DFE is equal to the angle GFE, and EDF to EGF and becaufe the angle DEF is equal to the angle GEF, and GEF to the angle ABC; therefore the angle ABC is equal to the angle DEF: For the fame reafon, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the fides, &c. Q.E. D.

:

84. I

IF

F two triangles have one angle of the one equal to one angle of the other, and the fides about the equal angles proportionals, the triangles fhall be equiangular, and fhall have thofe angles equal which are oppofite to the homologous fides.

Book VI.

a 23. 1.

Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the fides about thofe angles proportionals; that is, BA to AC, as ED to DF; the triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE.

At the points D, F, in the straight line DF, make the angle FDG equal to

either of the angles

BAC, EDF; and the
angle DFG equal
to the angle ACB;
wherefore the re-
maining angle at B
is equal to the re-

b 32. I. maining one at Gb,
and confequently the

B

A

triangle ABC is equiangular to the triangle DGF; and c4. 6. therefore BA: AC:: GD: DF. But by hypothefis, BA: AC: ED: DF; and therefore

d11.3. € 9. 5.

ED: DF:: GD: dDF; wherefore ED is equal e to DG: And DF is common to the two triangles EDF, GDF; therefore the two fides ED, DF are equal to the two fides GD, DF; but the angle EDF is alfo equal to the angle GDF; f 4.1. wherefore the bafe EF is equal to the bafe FG, and the triangle EDF to the triangle GDF, and the remaining angles to the remaining angles, each to each, which are fubtended by the equal fides: 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 ACB; therefore the angle ACB is equal to the angle DFE; and the angle BAC is equal to the angle EDF g; wherefore alfo the remaining angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D.

g Hyp.

PROP.

IF

F two triangles have one angle of the one equal to one angle of the other, and the fides about two other angles proportionals, then, if each of the remaining angles be either lefs, or not lefs, than a right angle, the triangles fhall be equiangular, and have thofe angles equal about which the fides are proportionals.

Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAC to the angle EDF, and the fides about two other angles ABC, DEF proportionals, fo that AB is to BC, as DE to EF; and in the first cafe, let each of the remaining angles at C, F, be lefs than a right angle. The triangle ABC is equiangular to the triangle DEF, that is, the angle ABC is equal to the angle DEF, and the remaining angle at C to the remaining angle at F.

For, if the angles ABC, DEF be not equal, one of them is greater than the other: Let ABC be the greater, and at the point B, in the straight line

Book VI.

AB, make the angle ABG equal to the angle a DEF:

A

And because the angle at

equal b to the remaining

b 32. 1.

angle DFE: Therefore the triangle ABG is equiangular to
the triangle DEF; wherefore, AB: BG:: DE: EF; but, c 4. 6.

by hypothefis, DE:EF::AB: BC,
therefore, AB: BC:: AB: BG";

and because AB has the fame ratio to each of the lines BC,

d 11. 5.

BG; BC is equale to BG, and therefore the angle BGC is e 9. 5. equal to the angle BCG f: But the angle BCG is, by hypo- f 6. 1. thefis, lefs than a right angle; therefore alfo the angle BGC

13. 1.

Book VI. is lefs than a right angle, and the adjacent angle AGB must be greater than a right angle . But it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle: But, by the hypothefis, it is lefs than a right angle; which is abfurd. Therefore the angles ABC, DEF are not unequal, that is, they are equal: And the angle at A is equal to the angle at D; wherefore the remaining angle at C is equal to the remaining angle at F: Therefore the triangle ABC is equiangular to the triangle DEF.

h 17. I.

Next, let each of the angles at C, F be not less than a right angle; the triangle ABC is alfo, in this cafe, equiangular to the triangle DEF.

The fame conftruction
being made, it may be
proved, in like manner,
that BC is equal to BG,
and the angle at C equal
to the angle BGC: But B
the angle at C is not lefs.

G

A

C E

D

than a right angle; therefore the angle BGC is not less than a right angle: Wherefore, two angles of the triangle BGC are together not lefs than two right angles, which is impoffible h; and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the firft cafe.

IN

PROP. VIII. THEOR.

Na right angled triangle, if a perpendicular be drawn from the right angle to the base; the triangles on each fide of it are fimilar to the whole triangle, and to one another.

Let ABC be a right angled triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base BC: the triangles ABD, ADC are fimilar to the whole triangle ABC, and to one another.

Because the angle BAC is equal to the angle ADB, each of them being a right angle, and the angle at B com

mon