Page images
PDF
EPUB

Let ABC, DEF, be two triangles, which have their sides proportionals, so that AB is to BC as DE to EF, and BC to CA as EF to FD, and, consequently, ex æquali, BA to AC as ED to DF: then the triangles ABC, DEF, shall be equiangular, and shall have those

[ocr errors]

F

angles equal which are opposite to the homologous sides, viz. the angle ABC to the angle DEF, and BCA to EFD, and BAC to EDF.

At the points E, F, in the straight line EF, make the angle FEG equal to the angle ABC, and the angle EFG to the angle BCA; therefore also the remaining angle EGF will be equal to the remaining angle BAC: Therefore the triangle ABC is equiangular to the triangle GEF, and, consequently (6. 4), they have their sides about the equal angles proportionals; therefore AB is to BC as GE to EF: But (Hyp.) AB is to BC as DE to EF; therefore (5. 11) DE is to EF as GE to EF, and therefore (5. 9) DE is equal to GE: And in like manner, DF is equal to GF: Therefore, because, in the two triangles DEF, GEF, the two sides DE, EF are equal to the two GE, EF, each to each, and the base DF to the base GF,-therefore the angle DEF is equal to the angle GEF, and the other angles to the other angles to which the equal sides are opposite, viz. the angle DFE to the angle GFE, and the angle EDF to the angle EGF: And because the angle DEF is equal to the angle GEF, and the angle GEF to the angle ABC, therefore also the angle ABC is equal to the angle DEF: And, for the like reason, 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 sides &c. Q. E.D.

S

PROP. VI. THEOR.

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 shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.

Let ABC, DEF be two triangles, which have the angle BAC of the one equal to the angle EDF of the other, and the sides about those angles proportionals, that is, BA to AC as ED to DF: then the triangles ABC, DEF shall be equiangular, and shall have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE.

B

CE F

G

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; therefore also the angle at B is equal to the angle at G, and the triangle ABC is equiangular to the triangle DGF: Therefore (6. 4) BA is to AC as GD to DF: But (Hyp.) BA is to AC as ED to DF; therefore GD is to DF as ED to DF, and therefore GD is equal to ED: Therefore, because in the two triangles GDF, EDF, the two sides GD, DF are equal to the two ED, DF, each to each, and the angle GDF to the angle EDF-therefore the base GF is equal to the base EF, and the triangle GDF to the triangle EDF, and the other angles to the other angles, each to each, to which the equal sides are opposite, viz. the angle DFG 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 (Hyp.); therefore also the angle at B is

equal to the angle at E, and the triangle ABC is equiangular to the triangle DEF.

Wherefore, If two triangles &c.

PROP. VII.

Q. E.D.

THEOR.

If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals, then, if each of the remaining angles be either less, or not less, than a right angle, or if one of them be a right angle, the triangles shall be equiangular, and shall have those angles equal about which the sides are proportionals. Let ABC, DEF be two triangles, which have one angle of the one equal to one angle of the other, viz. the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF, proportionals, viz. AB to BC as DE to EF; and, first, let each of the remaining angles at C, F be less than a right angle: then the triangles ABC, DEF shall be equiangular, and shall have the angle ABC equal to the angle DEF, and the angle at C to the angle at F. For, if the angles ABC, DEF be not equal, let ABC be the greater, and at the point B, in the straight line AB, make the angle ABG equal to the angle DEF: Then, because the angle BAG is equal to EDF, and ABG to DEF, therefore the angles AGB, DFE are equal, and the triangle ABG is equiangular to DEF: Therefore AB is to BG as DE to EF: But (Hyp.) AB is to BC as DE to EF; therefore AB is to BC as AB to BG, and therefore BC is equal to BG, and the angle BCG to the angle BGC: But BCG is less than a right angle; therefore also BGC is less than a right angle, and therefore AGB must be greater than a right angle: But the angle AGB was proved equal to DFE; therefore also the angle

[ocr errors]

B

DFE is greater than a right angle: But it is also less (Hyp.)-which is absurd: Therefore the angle ABC is not unequal to DEF, that is, it is equal to it, and the angle at C to the angle at F.

4.4

B

E

с

Next, let each of the angles at C, F be not less than a right angle: Then, if the angle ABC be not equal to DEF, let ABC be the greater, and make the angle ABG equal to DEF: Then it may be proved, as before, that BC is equal to BG, and the angle BCG to the angle BGC: But BCG is not less than a right angle; therefore also BGC is not less than a right angle, that is, two angles of the triangle BCG are together not less than two right angles-which is impossible (1. 17): Therefore the angle ABC is equal to the angle DEF, and the angle at C to the angle at F.

A

G

E

D

F

Lastly, let one of the angles at C, F, viz. the angle at C, be a right angle: Then, if the angle ABC be not equal to the angle DEF, it must be either greater or less; in either case, make the angle ABG equal to DEF: Then, as before, it may be shewn that the angle BCG is equal to the angle BGC: But B BCG is a right angle; therefore also BGC is a right angle, that is, two angles of the triangle BCG are together equal to two right angles-which is impossible: Therefore the angle ABC is equal to the angle DEF, and the angle at C to the angle at F. Wherefore, If two triangles &c. Q. E.D.

PROP. VIII. THEOR.

In a right-angled triangle, if a perpendicular be drawn from: the right angle to the base, the triangles on each side of it are similar 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, ACD shall be similar 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 that the angle at B is common to the two triangles ABC,

ABD, therefore the third angle ACB is
equal to the third angle BAD: There-
fore the triangles ABC, ABD are equi-B
angular, and have the sides about their

equal angles proportionals (6. 4), that is, they are similar (6. Def. 1): And, in like manner, it may be shewn that the triangles ABC, ACD are equiangular and similar: Therefore the triangles ABD, ACD, being both equiangular and similar to ABC, are equiangular and similar to each other.

Wherefore, In a right-angled triangle &c. Q.E.D.

COR. From this it is manifest, that the perpendicular, drawn from the right angle of a right-angled triangle to the base, is a mean proportional between the segments of the base, and also that each of the sides is a mean proportional between the base and its segment adjacent to that side: For, in the triangles ABD, ACD, BD is to DA as DA to DC, and in the triangles ABC, ABD, CB is to BA as BA to BD, and in the triangles ABC, ACD, BC is to CA as CA to CD.

PROP. IX. PROB.

From a given straight line to cut off any part required. Let AB be the given straight line: it is required to cut off any part required from it (5. Def. 1).

« PreviousContinue »