E. 1 | Hyp. At · B in AB, let BC and

BD make the adjacent
Zs ABC,ABD together

= two rt. angles;
2 Conc. then BC and BD will form

one and the same st.

line CD C. 1 by Sup. Should BD not be it the

same st. line with CB, Pst. 2. make BE in the same st. line with CB. D. 1 by Hyp. ::: AB with the st. line CBE makes the Zs ABC, ABE,

adjacent angles,
2 P. 13. these Zs ABC, ABE together = two rt. angles:

Нур. . But Zs ABC and ABD also = two rt. angles,
Ax. 1. .. LS ABC and ABE = Ls ABC and ABD:

Sub. taking away the common ABC, 6 Ax. 3. ... the rem. Z ABE= the rem. _ ABD, 7 ex. abs. and the less Z ABE the greater ABD; which is

impossible; 8 Conc. .. BE is not in the same st. line with CB. 9 Sim. In the same way no line except BD is in the same st. line

with CB; 10 Conc. .:. BD is in one and the same st. line with CB. 11 Rec. Wherefore if at a point in a line, &c.

Q.E.D. Sch.-The words “upon the opposite sides of it" are of essential importance ; for two lines may make with a third, two angles equal to two rt. angles, and yet the two lines not be in one st. line.

3 4 5


PROP. 15.-THEOR. If two st. lines cut one another the opposite, or verticalangles shall be equal.

DEM.–P: 13, Ax. 1, Ax, 3, Ax, 2. Е. І, Нур. . Let the two st. lines AB,

CD intersect in the · E; 2 Conc. then _ AEC = _ DEB, & L CEB = L AED.

D. 1 by Hyp. ..: AE makes with CD the

P. 13 .. these s CEA & AED

together=twort.angles. 3 Нур. . And :: DE makes with AB the Zs AED, DEB; 4 P. 13. . , these Zs AED and DEB together = two rt. angles;

Ax. 1. also Zs CEA, AED = US AED, DEB:

Sub. Take away the common angle AED, 7 Ax. 3. and rem. Z CEA will = rem. Z DEB.

Sim. In like manner we prove Z CEB = L'AED. 91 Rec. Therefore, if two st. lines cut one another, &c. Q.E.D.

Cor. 1.The angles formed by two st. lines, AB and CD, cutting each other in one point E, are together equal to four right angles.


5 6

D. 11 by P. 13. Zs AEC and CEB together = two rt. angles,

2 Pr, 13. and Zs AED and DEB also two rt. angles;
3 Add. .. by adding the equals to the equals,
4 Ax, 2. the Zs AEC, CEB, AED, and DEB together


four rt.

Cor. 2.- And all the angles formed by any number of st. lines, AC, BC,
DC, EC, FC, 8c., diverging from a common centre, C, are together equal to
four rt. angles,
C. 1jby Pst. 1. Produce any two of the B

given lines, as AC to G,
BC to H.

D. 1 by Cor. 1,15 ::: AG and BH intersect
in the point C,

G .. the Zs ACB, BCG, ACH and HCG toge

ther = four rt. Zs. 2 P. 13. But _s ACB, BCD, DCG, together = US F

E ACB and BCG;

HI 31 P. 13. also Zs ACF, FCE, and ECG, together = LS ACH and

Con. .. all the angles diverging from · C are together equal to

four rt. angles. Sch.-1. This proposition might be briefly proved by saying that the opposite angles are equal, because they have the same supplement.

2. “ This Prop. is the developement of the definition of an angle. If the lines at the angular point be produced, the produced lines have the same inclination to one another as the original lines have.”-Potts' Euclid, p. 49.

3. The converse of this Prop. is, “ If foúr lines meeting in a point make the vertical angles equal, each alternate pair of lines shall be in one and the same st. line."

APP.-To find the breadth of a lake, or the distance between two inaccessible objects, A and B.


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If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite ungles.

Con.-P. 10, P. 3, Pst. 1, Pst. 2.-DEM.-P. 15, P. 4, As. 9.


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E. 11 Нур.
Let BC, a side of a A

ABC, be produced

to D;
2 Conc.

then the ext. / ACD
> L CBA, or than

_ CAB, the int. / s.
C. 1 by P. 10 & Pst. 1 Make CE = AE,

and join BE;
2 Pst. 2, P. 3,1 produce BE to? F,

Pst. 1. so that BE = EF,

and join FC.
D. ) by C. 1 & 2, In AS ABE, CEF,

..: AE

and BE = EF, 2 P. 15.

and L AEB= L CEF ; 3 P 4.

..Δ ΑΕΒ A CEF, and L. BAE
4 C.

But Z ECD or ACD > L ECF;
Ax. 9.

and .. LACD > _BAC.
Sim. P. 10. In like manner produce AC to G, and bisect BC,
D. 1-5. and / BCG, i.e., L ACD, will be > 7. ABC.
Rec. Therefore if one side of a triangle be produced, &c. Q.E.D.

Prop. 17.—THEOR. (converse of Ax. 12.)
Any tuo angles of a triangle are together less than two right angles.

Con.-Pst. 2, DEM.-P. 16, Ax. 4, P. 13.
E. 11 Нур. .

Let ABC be a triangle ;
2 Conc. then any two Zs in it, as A

and B, are together < two rt.

angles. C. 1 by Pst. 2. Produce BC to D. D. 1 by Con. :: LACD is the exterior angle,


D 2 P 16... LACD > LB or LA. 3 Add. To each of these unequals add

the L ACB Ax. 4. then Z8 ACD and ACB > Ls ABC and ACB. 5 P. 13. But Zs ACD and ACB two rt. angles.

D. 4. .. Zs ABC, ACB < two rt. angles. 7 Sim. In like manner Zs BAC, ACB < two rt, angles; 8 Sim. and ZS BAC, ABC < two rt. angles. 91

Rec. Therefore, any two angles of a triangle, &c. Q.E.D. Sca. Both the 16th and the 17th propositions will be included in the thirty-second.

The greater side of every triangle is opposite to the greater angle.

Con.-P. 3, ?st. 1.-DEM.-P. 5, P. 16.
E. 1| Hyp.

In A ABC, let AC be greater

than AB; 2 Conc. then _ ABC is > L ACB. C. 1 byP.3&Pst. 1 AC make AD =

AB, and join BD. D. 1 by C. & P. 5. In A ABD, ::: AD

1... Z ADB


- AB,

2 P. 16. But ext. ADB > int. _ DCB,
3 D. 1. and ADB L. ABD;
4 D. 2, 3. . . _ ABD > L DCB;
5 à fort.

much more is _ ABC > L ACB. 6 Rec. Therefore, the greater side of every triangle, &c. Q.E.D.

ScH.-The argument on which the conclusion depends is named “a fortiori,by the stronger reason, and proves that the given predicate belongs in a greater degree to one subject than to another; as in the Syllogism,-Y is greater than Z, and X greater than Y; much more is X greater than Z.

PROP. 19.-THEOR. (Converse of the 18th.) The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

DEM.-P. 5. P. 18.

E. 1 Hyp.

In A ABC let ABC be >

then side AC is > side


2 Conc.


SUP.- The side AC must be greater than, equal to, or less than AB. D. 1,by Sup. & P.5. If AC = AB, then _ ABC = L. ACB: 2 Hyp.

but _ABC + L ACB; 3 D. 1 & 2. .. AC + AB. 4 Sup. & P. 18. Again, if AC is < AB, then _ ABC is< L ACB: 5 Hyp.

but ABC is * L ACB; 61 D. 4 & 5. :: AC is * AB. 7 D. 3 & 6. Now, AC is neither equal to, nor less than, AB; 8 Concl.

AC is greater than AB. 9 Rec. Wherefore, the greater angle of every triangle, gc. Q.E.D.

SCH.-1. Props. 5, 6, 18, and 19, may be combined into one thus,—“ One angle of a triangle is greater than, equal to, or less than, another angle, as the side opposed to the one is greater than, equal to, or less than, the side opposed to the other; and vice versa."

2. By Props. 17, 18, and 19, we may, frove that from the same point there can be drawn but one perpendicular to a given line, and that this perpendicular is the shortest of all the lines from the given point to the given line.

3. As from a given point only three lines can be drawn perpendicular to each other, it is impossible to imagine that there are more than three species of quantity,--a Line a Surface, and a Solid.

Any two sides of a triangle are together greater than the third side.

Con.-Pst. 2, P. 3, Pst. 1.-DEM.P.5, Ax. 9, P. 19.
E. 1) Hyp. Let ABC be a triangle;
2 Conc. 1. then the sides AB and AC to-

gether are > BC;
2. sides AB and BC > AC;

BC and CA > AB.
C. il by Pst. 2. Produce BA,
2 P. 3. and make AD AC;

Pst. 1. join D and C. D. I by C. 1 &P.5.1AD= AC,



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D. 2 C. 1 & 2 but _ BCD is > LACD ;

3 Ax. 9 .. BCD is > LADC or BDC ; 4 Ax. 9. and in a DBC

Z BCD is > LBDC, 5 P. 19. ... also BD is > BC. 6 C. 1. But BD BA and AC together ; 7 D. 4 & 5. . . BA and AC together are > BC. 8 Sim. In like manner it may be proved that AB and BC are

> AC, 9

and BC and AC > AB. 10' Conc. & R. Therefore any two sides of a triangle, &c. Q.E.D.

OR, C.

by P. 9. Let AE bisect _ BAC. D. 1 by P. 16. ::: _ BEA is > _ EAC, and L CEA than EAB ; 2 P. 19.

BA is > BE, and AC than CE : 3 Conc. And BA + AC > BE + CE, which are equal to BC. 4 Rec. Therefore any two sides, &c.

Q.E.D. COR.The difference of any two sides of a triangle is less than the rem. side. D. 11 by P. 20. The sides AC and BC are > AB ;

2 Sub. take AC both from (AC + BC,) and also from AB; 3 Ax. 5. .. BC is > AB diminished by AC, or than the difference

between AB and AC. APP.-1. Of all lines that can be drawn from one point, A, to another I, and reflected to a third point, B, those are shortest, AE, EB, which make the angle of incidence, AEG, equal to the angle of reflection, BED. C. 11 by P. 12 & Pst. 2. From B draw BD

perp. to DG,

B and produce

BDindefinitely; 2 P. 3.

make DC=DB, D

and join E, C. 3) Assum., Pst. 1. Also in DG as

sume another
point F, & join

AF, BF, and CF.
D. I by C. 1, 2. In triangles BED, CED, DE is common, BD equal

to DC, and the angles at D equal, 2 P. 4.

BE is = CE, and L. BED = L CED. 3) Sim.

In the same way we prove BF to be CF. 4 H. & C. & D. 2 Now _s BED, DEC, and AEG are all equal ; 5 C.

and AEC is a st. line: 6 D. 5 & C. 3, P. 20 | Also AFC is a A; ., AF and FC > AC: 71 D. 2, 3. but AC = AE and EB, and FB :

= FC; 8 Conc.

.:. AF, FC, or AF, FB, > AE and EB. 9) Rec.

Wherefore, of all lines that can be drawn &c. Q.E.D. 2. Natural causes act by the shortest lines; therefore all reflections are made by the lines which cause the angle of reflection to equal the angle of incidence. Hence, by means of a mirror placed horizontally, we may construct a triangle the perpendicular of which shall be representative of the height of any object.-See“ Euclid Practically Applied," Part I., Pr. 20, 1. p. 84.

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