point E.

PROP. 21.-THEOR. If from the ends of the side of a triangle there be drawn two st. lines to a point within the triangle, these lines shall be less than the other two sides of the triangle, but shall contain a greater angle.

Con.-Pst. 2. DEM.-P. 20. Ax. 4. P. 16. E. 1 Hyp.

Let the a be ABC, and from B

and C let the st. lines BD, CD,
meet within the A in the · D;

2 Conc. 1. then BD and CD shall be < BA

and CA;
2. but the Ľ BDC shall be > than

< BAC.
by Pst. 2. Produce BD to meet CA in the

D. 1 by P. 20. In A ABE the two sides AB and

AE are > BE; 2 Add. to each of these add the line EC ; 3 Ax. 4. then the lines AB, AE, and EC, are > BE and EC. 4 P. 20. Again in A CED, the two sides CE and ED are > CD; 5 Add. to each of these add the line DB; 6 Ax. 4. then CE, ED, and DB together are > CD and DB together. 7 D. 3. But AB and AC are > BE and EC ; 81 à fort.

much more are AB and AC > CD and DB. 9 P. 16. Again, in A CDE the ext. L BDC is > the int. CED, 10 P. 16. and in A ABE the ext. L CED is > the int. _ BAC; 11 à fort.

much more is Z BDC > L BAC. 12' Rec. Therefore, if from the ends of a side, &c.

Q.E.D. APP.-In Optics this proposition is used to prove, that if from A we could see the line BC, and also from D a point nearer to the line, the base BC would appear less from A than trom D: it does this on the principle that quantities seen under a greater angle appear greater. For this reason the apparent diameter of the sun measures more when the earth sin perihelion, than when it is in aphelion. Andthus,--according to Vitruvius,who composed his work on Architecture, about 15 B.C.-the tops of very high pillars should be made but little tapering, because they will, from the distance, of themselves seem less.

PROP. 22.-PROB. To make a triangle, of which the sides shall be equal to three given st. lines, but any two whatever of these must be greater than the third.

SOL.-P. 20, P. 3, Pst. 3, Pst. 1, Def. 21.-DEM.-Def. 15, Ax. 1. E. 1, Data. Given the thrce st. lines A, B, and C, any two of which

are greater than the third, P. 20. A and B > C, A and C > B, and B and C >A; 2 Quæs. to make a triangle with sides = A, B, and C, each to each. C. 1 by Assum. . Take a st. line DE unlimited towards E;

2 P. 3. make ĐF= A, FG = B, and GH = C;
3 Pst. 3. from centre F, with DF, describe the ODKL,

3. and from centre G, with GH, the O HLK;

1. join the points F,K, and G,K; Def. 21. then the fig. KFG is a A, having FK = A, FG = B, Sol.

and GK = C.

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D. 1 byC.3,Def.15 ::: F is the centre of ODKL, FK = FD;

2 C. 2 & Ax. 1/ but FD = the st. line A, .. FK also = A.
3 C.3, Def. 15 Again, ::: G is the centre of O HKL, .. GH = GK;
+ C. 2 & Ax.1 but GH = C;... also GK=C;
5 C. 2. and FG is equal to B;
6 D. 2, 4. & 5., the three st. lines FK, FG, GK are respectively equal

to the three st. lines A, B, and C. 7 Rec. And therefore the A FKG has its three sides, &c. Q.E.F. Sca.-1. In P. 22 it is assumed that the two circles will have at least one point of intersection.

8. If two of the given lines were together equal to the third, the circles would touchi externally; if the two were together less than the third, the circles would not touch at all: in either case no triangle could be drawn.

APP.-1. All rectil. figures being divisible into triangles, this Prop. is of very extensive use-either for making one rectil. figure equal to another, or on the theory of Representative Values, making one figure like to another : in the first case the triangles into which the rectil. figure has been divided are repeated, side for side, in another rectil. figure of exactly the same linear dimensions: and the construction being completed, the two would correspond, angle to angle, line to line, and point to point: in the second case, the sides and angles of the first must be measured, and from a scale of equal parts, lines drawn in the second, representative of those in the first, and angles in the second equal to those in the first ;--for equality of angles, according to Def. 1, bk. vi., is essential to similarity of figure.

PROP. 23.-PROB. At a given point in a given line to make a rectilineal angle equal to a given rectilineal

Sol.-Pst. 1, P. 22.- DEM.-P. 8.


А Given the point A in

the st, line A B, and

the rectil. _ DCE; 2 Quæs.

required at in AB

to make an LEDCE. C. 1 by Assum In CD, CE take points

Pst 1. D & E, & join D,E;

with AF=CD, AG=
CE, & FG = DE,

make the A AFG ; D
33. then the _ FAG shall be
Dil by C. 2. FA=DC, AG =CE,&

2 P. 8. .. he L. FAG= _ DCE.
3 Rec.
Therefore, at point A in AB &c


E. 1 Dat.





Sch.--This Proposition is an extension, or rather a generalization, of the eleventh : by the eleventh we draw an angle of a particular species,-a right angle; but by the twentythird we draw any angle whatever.

APP.-1. Pr. 23 is of the widest use, in Surveying, Engineering, Perspective, and indeed in all the other parts of Practical Mathematics.

2. To Construct a Line of Chords for measuring and making angles of a determinate magnitude.

Take a circle, and draw a diameter
DA; bisect DA in C, and at C draw CB

at rt, angles to DA, and produce BC to
E;—the circle will be divided into four
quadrants, each 90°. Let the arc AB of
909 be divided into arcs of 109, 209 300
&c., up to 90. Draw AB, the chord of
900, and from A, as a centre, set off on
AB the chords of 109, 200, 30?, &c.: the
divisions on the line AB will be a line of

A N.B. The chord of 609 is equal to

с the radius of the circle.

In triangle DCF, let the angle DCF measure 60°,-then, since CF and CD are equal, the angles CDF and CFD will be equal to each other, each measuring 609. The triangle, being equiang. is also equil.; and DF, the chord of 60%, equals the radius DC.

On this property the use of a Line of Chords depends.

PROP. 24.-THEOR. If two triangles have two sides of the one equal to the two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides of the other ; the base of that which has the greater angle shall be greater than the base of the other.

Con.-P 23, P. 3, Pst. 1.-DEM ,-P.4, P. 5, Ax. 9, P. 19. E. 11 Hyp. 1.

In the two AS

let AB= DE

& AC=DF;
2. but _ BAC be

3 Conc. then the base BC

is>thebase EF. C. 1 by Hyp.&P.23 LetDFbe{DE, & at D in DE

H н make LEDG B

L BAC; 2 P. 3, & Pst. 1 also make DG AC or DF, and join E, G ; G,F. :D. 1 by Hyp. & C.1

T: AB = DE, AC, DG, & L BAC = LEDG, 2 P. 4.

base EG is equal to base BC. 3 C. 2 & P. 5. | And ::: DG= DF, . . L DGF = [ DFG. 41 Ax. 9, But _ DGF is > EGF, and DFG > LEGF; 51 à fort

much more is _ EFG > LEGF. D. 5.

And A EFG, the EFG > the _ EGF, 71 P. 19 & the greater _ is opposite to the greater side ; 8.D. 5 & P. 19 ... the side EG is > side EF. 9) D. 2. But EG = BC, and .:. BC is > EF. 101 Rec.

Therefore, if two triangles have two sides, &c. Q.E.D.

Scu.-In Pr. 24 it is assumed that D and F will be on different sider of EG; or in other words that DH is less than DF or DG.


PROP. 25,-THEOR If two triangles have two sides of the one equal to the two sides of the other, each to each, but the base of one greater than the base of the other ; the angle contained by the sides of the one which has the greater base shall be greater than the angle contained by the sides equal to them of the other,

DEM.-P. 4, P. 24. E. 1 Hyp. 1 In AS ABC, DEF, let AB DE, &

A AC= DF, 2 but let BC be > EF, 3 Conc. then the BAC is

> the _ EDF. D. 1 Sup. 1. If / BAC be L

EDF, it is either

equal or less. 2 Sup. 2. Suppose that _BAC


E F is = _ EDF: 3 Hyp. & P.4 then base BC = base EF : 4 Нур. . But BC is + EF; 5 Conc. .: 4. BAC is # Z EDF. 6 Sup. Again, suppose _ BAC< L EDF ; 7 H. & P. 24 then base BC is <base EF : 8! Нур. . But BC is * EF ; 91 Conc. < BAC is * _ EDF. 10 D, 5 & 9. Now, _ BAC is #, nor <, < EDF ; 11 Conc.

< BAC is > Z EDF. 12 Rec.

Therefore, if two triangles have two sides, &c. Q.E.F. SCH.—Prop. 24 and 25 have the same relation to each other as Props. 4 and 8, and the four may be combined thus:- Iftwo triangles have two sides of the one respectively equal to tuo sides of the other, the rem. side of the one will be greater or less than, or equalto, the rem. side of the other, according as the angle opposed to it in the one is greater or less than, or equal to, the angle opposed to it in the other; or vice versa.–LARDNER'S Euclid, p. 56.

PROP. 26.--THEOR.—(Important.) If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz., either the sides adjacent to the equal angles in each, or the sides opposite to them, then shall the other sides be equal, each to each, and also the third angle of the one to the third angle of the other.

Con.-P. 3, Pst. 1.-DEM.-P. 4, Ax 1, P. 16. E. 1 Hyp. 1. In As ABC, DEF, let / ABC = _ DEF, and ACB

= L DFE;
2 2. also let one side :

= one side ;
3 Conc. 1. then the other sides shall the other sides;

2. and the third L of the one = the third Z. of the other.

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Case. I.-Let the equal sides be adjacent to the equal angles. E. 1 Hyp. 3. Let side BC= side EF;

2 Conc. 1. then AB =
DE, and

2. and / BAC=_EDF.

C. 1 by Sup. 1. For if AB is # DE,

one is > the other, 2 Sup. 2. Let AB be > DE: 3 P.3 & Pst. 1 Make BG = DE, and

join G,C. 1). 1 by C & H.3 :: in As GBC, DEF,


ED, BC =

EF, and _ GBC = _ DEF;
2 P. 4. .:. base GC = base DF, A GBC = A DEF,

and / BCG = _ DFE : 3 Hyp. 1. But L DFE = l BCA ; 41 Ax. 1. .. BCG = _ BCA, the less 2 equals the greater Z

which is impossible. 5 Conc.

AB is not # DE, i.e., AB does = DE ; 6 D.5 & H. 3. Hence in As ABC, DEF, :: AB = DE, BC = EF, H. 1.

and _ ABC, = _ DEF ; 7 P. 4.

base AC = base DF, and third _ BAC=third EDF.

BG =

CASE II.-Let the equal sides be opposite the equal angles in each.
E. 1| Hyp. 4.

Let AB DE, ZB=
L E, and _ ACB


D LDFE; 2 Conc. 1. then AC shall

EF, and

DF, BC= EF, 3 2. and third BAC

third _ EDF. C. 1 by Sup. 3. For if BC# EF, one of

them is the greater,
2 Sup. 4. Let BC be > EF ;
3 P.3, Pst. 1 Make BH


F join A,H. 1). 1 byC.3 & H.4 :: in As ABH, DEF, BH = EF, AB = DE, and L

ABH = L DEF ; 2 P. 4. .. base AH = DF, A ABH A DEF,

and _ BHA Z EFD ; 31 Hyp. 1 But / EFD _ BCA ; 41 Ax. 1 .. _ BCA = L BHA, or the int.

the ext. angle ; 5) P. 16. which is impossible : 6 Conc. BC is not #EF, i.e., BC = 7 H, 4 & D.6 Hence in As ABC, DEF, : : AB = DE, BC = EF, and H. 1.


- L DEF; 8 P. 4. .:. base AC=base DF, and third _ BAC= third _EDF. 9 Rec. Wherefore, if two triangles have two angles of the one, &c.



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