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PROP. 22. PROB.

H

To make a triangle of which the sides shall be equal to three given straight lines, any two of which are together greater than the third. Given three lines A, B,

and C, any two of
which are together

I.
> the third.

II. To make a A of which the sides shall be

E.

Н. equal to A, B, and

C, each to each. Take DE, a st. line of

unlimited length. Cut off DF = A, FG =B, GH=C (Post. 3). With F as centre and FD as rad., describe O I.; and with G as centre and GH as rad., describe O II.,

cutting O I. at K. Join KF, KG. :F is centre of O I., FK = FD (Def. 11). But FD=A (cons.), : FK = A. Similarly, GK = GH=C, and FG=B (cons.), :: AFGK has its sides respectively equal to A, B,

and C. Q.E.F. 54. Show in the above that the circles must cut one another, and that we can make two triangles satisfying the given conditions.

55. Show by diagrams what form the figure would take if one of the given lines be taken (1) greater than, (2) equal to, the other two together.

56. Construct a rhombus on a given line, one of the diagonals of which shall be equal to the given line.

57. To construct a right-angled triangle, (1) with one side and the hypotenuse equal respectively to two given lines ; (2) with sides equal respectively to two given lines.

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PROP. 23. PROB. At a given point in a given straight line, to make an angle equal to a given angle, i.e., so that the given line shall be one of the arms of the angle.

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LC.

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: AF

:: LA

Given line AB with point A in it, and < C.
To make at A in AB an angle
Take D and E, any points in CD, CE, and join them.
Cut off from AB, AF CD (Post. 3), and from A as

centre, with CE as rad., describe a O; from F as
centre, with DE as rad., describe a O, cutting the
first o at G (Post. 3). Join AG and FG.

CD, AG = CE, and FG = DE (cons.), ::AFAG ADCE (1. 8).

LC. :: at A in AB an has been described = L C.

Q.E.F. Note.-In this proposition, after joining DE, Euclid says," Make a triangle AFG whose sides shall be equal to CD, DE, and EC (1. 22)," and then proof follows as above by 1. 8. It seems better to make our construction more detailed, following the method of 1. 22, that we may be certain of the position of the constructed triangle.

58. To make a triangle whose sides shall be respectively equal to the halves of the sides of a given triangle ; or to the doubles of the sides of a given triangle.

59. To construct a triangle having given the base and the two angles at the base, which are together less than two right angles (I. 17).

60. To construct a triangle whose base shall be equal to a given line, one of the angles at the base equal to a given angle, and the sum of the other two sides equal to a given line.

a

For Euclid I. 24, see Appendix.

PROP. 24. THEOR. If two triangles have two sides of the one equal to 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 equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other.

D

А

G

H

: AC

Given triangles ABC and DEF, having AB = DE,

AC DF, and LA > _EDF. To prove BC > EF. Apply ABC to A DEF, so that A may be on D, and AC along DF, then C will fall on F,

DF (given), and let A ABC take the position of DGF

on the side of DF remote from A DEF; Then DG is AB, FG is BC, and LGDF is 2 A. Bisect – EDG, and bisector must fall within - FDG,

: ¿FDG > LEDF; let it meet FG at H; join EH. In triangles EDH, GDH, :: ED = DG, DH is common, and _ EDH < GDH (cons.),

HG (I. 4). To each add HF. .. EH + HF FG, but EH + HF > EF (I. 20).

:: FG, i.e., BC > EF. Q.E.D.

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:: EH

PROP. 25. THEOR.

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

E

B

Given AB DE, AC = DF, and BC >EF.
To prove LA > D.
If A is not > LD, it must be either = or <LD.
LA is not = _D, for then BC must be

EF (I. 4), whicb it is not. Nor is LA < _ D, for then BC must be < EF (I. 24),

which it is not. :: L A is neither = nor < LD; :: LA is> D. Q.E.D. 61. Every straight line drawn from the vertex of a triangle to the base is less than the greater of the other two sides, or than either of them if they be equal.

62. In I. 24, let AB be not greater than AC, and let A DEF be superposed on ABC, so that DE coincides with AB ; then prove point F falls on the other side of BC from A.

63. The straight lines drawn to each of two given points on opposite sides of a given line, from any point in the given line, are equal ; prove that the line joining the two points will be bisected by the given line at right angles.

с

PROP. 26. THEOR. 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, namely, either (1st) the side adjacent to the equal angles in each, or 2nd) a side opposite to one of the equal angles in each, then the two triangles shall be equal in every respect.

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B

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1. Given triangles ABC and DEF, having

_ B = LE, 2C = _ F, and BC EF. To prove BA ED, AC = DF, and 2A = _ D. Place A ABC on A DEF, so that B is on E, and BC

along EF, then C falls on F, :: BC = EF; and BA must lie along ED, :: _B = LE; and CA must lie along FD, :: LC :: pt. A must fall on D, and A ABC must coincide with A DEF.

:: BA = ED, AC = DF, and LA = _D.

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2. Given triangles ABC and DEF, having

= LE, LC = _F, and AB DE. To prove BC EF, AC = DF, and - A = _D.

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