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equal Angles, are homologoris, or of like Ratio; which was to be demonstrated.

PROPOSITION V.

23.

THEOREM.
If the Sides of two Triangles are proportional, the

Triangles Mall be equiangular ; and their Angles,
under which the homologous Sides are fubtended,

are equal. I

ET there be two Triangles, ABC, DE F, hav

ing their Sides proportional, that is, let AB be to BC, as D E is to EF; and as BC to CA, so is EF to FD. And also as BA to CA, fo ED to DF, I say, the Triangle ABC is equiangular to the Triangle DEF; and the Angles equal, under which the homologous Şides are subtended, viz. the" Angle ABC, equal to the Angle DEF; and the Angle BCA equal to the Angle EFD, and the Angle BAC equal to the Angle EDF.

For at the Points E and F, with the Line EF, make * the Angle FEG, equal to the Angle AB C; and

the Angle EFG, equal to the Angle BCA: Then + Cor. 32. I. the remaining Angle BAC, is f equal to the remain

ing Angle EGF.

And so the Triangle ABC is equiangular to the Triangle EGF; and consequently the sides that are

fubtended under the equal Angles, are proportional. 14 of this. Therefore as AB is to BC, so is I GE to EF; but

as AB is to BC, so is DE to EF: Therefore as DE is to EF, fo is * GE to EF. And since DE, EG, have the same Proportion to EF, DE shall be + equal to E G. For the same Reason, DF is equal to FG; but EF is common. Then because the two Sides DE, EF, are equal to the two Sides GE, EF, and

the Base DF is equal to the Base F G, the Angle 18. 3. DEF is equal to the Angle GEF; and the Trian

gle DEF equal to the Triangle GEF; and the other Angles of the one, equal to the other Angles of the other, which are fubtended by the equal Sidesa Therefore the Angle DEF is equal to the Angle GEF, and the Angle EDF equal to the Angle

EGF;

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EGF; and because the Angle DEF is equal to the Angle GEF; and the Angle GEF equal to the Angle ABC; therefore the Angle ABC shall be also equal to the Angle FED; For the same Reason, the Angle ACB shall be equal to the Angle DF E; as also the Angle A equal to the Angle D; therefore the Triangle ABC will be equiangular to the Triangle DEF. Wherefore, if the Sides of two Triangles are proportional, the Triangles shall be equiangular ; and their Angles, under which the homologous Sides are subtended, are equal; which was to be den monstrated.

PROPOSITION VI.

TH E O R E M. If two Triangles have one Angle, of the one equal to one Angle of the other ; änd

if the Sides about the equal Angles be proportional, then the Triangles are equiangular, and have those Angles equal, under which are subtended the homologous Sides. L

ET there be two Triangles ABC, DEF, having EDF of the other; and let the sides about the equal Angles be proportional, viz. let AB be to AC, as ED is to DF. I say, the Triangle ABC is equiangular to the Triangle DEF; and the Angle ABC equal to the Angle DEF; and the Angle ACB equal to the Angle DF E.

For at the Points D and F, with the Right Line DF, make * the Angle FDG equal to either of the * 23. I. Angles BAC, EDF; and the Angle D F G equal to the Angle ACB.

Then the other Angle B, is † equal to the other † Cor. 32. 1. Angle G; and so the Triangle ABC, is equiangular to the Triangle DGF; and consequently, as B A is to AC, so is I G D to DF: But (by the Hyp.) as I 4 of tbis. BA is to AC, fo is ED to DF. Therefore as ED is * to DF, so is GD to DF; whence ED is † equal * 11. 5. to DG, and DF is common; therefore the two sides 9. 5. ED, DF, are equal to the two Sides GD, DF; and the Angle EDF, equal to the Angle GDF:

Con

Lgle of the one, equal to one Angle of the other,

Consequently the Base EF is * equal to the Bafe FG, and the Triangle DEF equal to the Triangle GDF; and the other Angles of the one, equal to the other Angles of the other, each to each; under which the equal Sides are fubtended. Therefore the Angle DFG is equal to the Angle DF E, and the Angle G, equal to the Angle E; but the Angle DF G, is equal to the Angle ACB: Wherefore the Angle

ACB is equal to the Angle DFE; and the Angle 1 Bs Flyp. BAC is falso equal to the Angle EDF: Therefore

the other Angle at B is equal to the other Angle at E; and so the Triangle ABC is equiangular to the Triangle D EF. Therefore, if two Triangles have one Angle of the one, equal to one Angle of the other; and if the Sides about the equal Angles be proportional, then the Triangles are equiangular; and have those Angles equal, under which are fubtended the homologouts Sides; which was to be demonstrate PROPOSITION VII,

THEO R E M. if there are two Triangles, having one Angle of the

one equal to one Angle of the other, and the Sides about other Angles

proportionat ; and if the remaining third Angles are either both less, or both not less than Right Angles, then all the Triangles be equiangular and have those Angles equal, about which are the proportional Sides.

ET two Triangles ABC, DEF, have one Anviz, the Angle BAC equal to the Angle EDF; and let the Sides about the other Angles ABC, DEF, Þe proportional; viz. as DE is to EF, so let AB be to BC; and let the other Angles at C and F, be both less, or both not less than Right Angles. I fay, the Triangle ABC is equiangular to the Triangle DEF; and the Angle A B C is equal to the Angle DEF, as also the other Angle at C, equal to the other Angle at F.

For if the Angle ABC be not equal to the Angle PEF, one of them will be the greater, which let be

ABC.

1

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ABC. Then at the Point B, with the Right Line
AB, make * the Angle AB G equal to the Angle* 23. I.
DEF.

Now because the Angle A is equal to the Angle
D, and the Angle A BG, equal to the Angle DEF;

the remaining Angle AGB, is equal to the remain- + Cor. 32, 1, Gi ing Angle DFE: And therefore the Triangle ABG,

is equiangular to the Triangle DEF; and so as AB
is to B G, fo is IDE to EF; but as DE is to EF, 4 of this,
fo is * AB to BC. Therefore as AB is to BC, so is * By Hype
AB to BG; and since AB has the same Proportion
to BC, that it has to BG, BC shall be + equal to 1 9: 5-
BG; and consequently the Angle at C equal to the
Angle BGC. Wherefore each of the Angles BCG,
or BGC is less than a Right Angle; and conse-
quently, AGB is greater than a Right Angle. But
the Angle AGB has been proved equal to the Angle
at F; therefore the Angle at F, is greater than a
Right Angle: But by the Hyp.) it is not greater,
fince C is not greater than a Right Angle, which is
absurd. Wherefore the Angle ABC is not unequal
to the Angle DEF; and so it must be equal to the
same; but the Angle at A is equal to that at D;
wherefore the Angle remaining at C is equal to the
remaining Angle at F; and consequently the Triangle
ABC is equiangular to the Triangle DEF. There-
fore, if there are two Triangles having one Angle of the
one, equal to one Angle of the other, and the sides about
other Angles proportional; and if the remaining third
Angles are either both less, or both not less than Right
Angles, then shall the Triangles be equiangular; and
bave those Angles equal, about which are the proportional
Sides; which was to be demonstrated,

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PROPOSITION VIII.

THEOREM.
If a Perpendicular be drawn, in a Right-lined

Triangle, from the Right Angle to the Base,
then the Triangles on each side of the Perpen-
dicular are similar both to the whole, and also
to one another.

ET ABC be a Right-angled Triangle, whose

Right Angle is BAC, and let the Perpendicular AD be drawn from the Point A to the Base B C. I say, the Triangles ABD, ADC, are similar to one another, and to the whole Triangle A B C.

For because the Angle BAC is equal to the Angle ADB, for each of them is a Right Angle, and the

Angle at B is common to the two Triangles A B C, * Cor. 32. 1. ABD, the remaining Angle ACB (hall be * equal to

the remaining Angle BAD. Therefore the. Triangle

ABC is equiangular to the Triangle A BD; and so * 4 of this as † BC, which subtends the Right Angle of the Tri

angle ABC, is to BA, fubtending the Right Angle of the Triangle ABD, so is AB fubtending the Angle C of the Triangle ABC to DB, fubtending an Angle equal to the Angle C, viz. the Angle BAD, of the Triangle ABD. And so moreover is A C to AD, subtending the Angle B, which is common to

the two Triangles. Therefore the Triangle ABC 1 Def. 1 of is fequiangular to the Triangle ABD; and the Sides ibis.

about the equal Angles are proportional. Wherefore the Triangle ABC is similar to the Triangle ABD. By the same way we demonstrate, that the Triangle ADC is also similar to the Triangle ABC. Wherefore each of the Triangles ABD, ADC, is similar to the whole Triangle.

I say, the said Triangles are also similar to one another.

For because the Right Angle B D A is equal to the Right Angle ADC, and the Angle BAD has been proved equal to the Angle C; it follows, that the reinaining Angle at B shall be equal to the remaining Angle DAČ. And so the Triangle ABD is equi

angular

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