PROP. XXI. THEOR.

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

Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle: BD and DC shall be less than the two sides BA, AC, but shall contain an angle BDC greater than the angle BAC.

Produce BD to E; and because two sides of a tri*20. 1. angle are greater than the third side, the two sides

BA, AE, of the triangle ABE are greater than BE. To each of these add EC; therefore the sides BA, 14. Ax. AC are greater than BE, EC.

+20. 1.

Again, because the two sides CE, ED of the triangle CED are greatert than CD, add DB to each of these; +4 Ax. therefore the sides CE, EB, are greater than CD, DB: but it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC, greater than BD, DC.

B

D

Again, because the exterior angle of a triangle #16. 1. is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC: and it has been demonstrated that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends, &c.

Q. E. D.

greater than BC, from each of these take AC," an operation which is impossible when AC exceeds BC. That the inference is, however, true, is easily shewn either by proving the absurdity of a contrary supposition, or by a direct demonstration.

PROP. XXII. PROB.

To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third.*

*20. 1.

Let A, B, C, be the three given straight lines, of which any two whatever are greater than the third, viz. A and B greater than C; A and C greater than B; and B and C greater than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each.

Take a straight line DE terminated at the point D, but unlimited towards E, and

make* DF equal to A,

*3. 1. FG equal to B, and GH

equal to C: from the centre F, at the distance FD, describe*

*3 Post.

the circle DKL ; and from

the centre G, at the distance GH, describe another circle

*3 Post.

HLK. Now FH, being equal to B and C together is greater than FD or FP, which is equal only to A; hence the point H in the circle LHK is farther from F than the point P in the circle DKL; the point H is therefore without the circle DKL, and, consequently, the circle LHK cannot be wholly within the circle DKL. Again, GD, being equal to B and A together, is greater than GH or GQ, which is equal only to C; hence the point D in the circle DKL is farther from G than the point Q in the circle LHK; the point D, therefore, is without the circle LHK, and, consequently, the circle DKL cannot be wholly within the circle LHK. Neither can they be wholly without each other; for the line, joining the centres of two circles wholly without each other, cannot be less than the two parts of it, from the centre of each circle to its circumference; whereas FG, the line joining the centres, is less than the two lines FD, GH, from the centre of each circle to its circumference. The circles must therefore be partly

within and partly without each other; they must, therefore, intersect. Let them intersect in K, and draw KF, KG: the triangle KFG shall have its sides equal to the three straight lines A, B, C, respectively.

*12 Def.

+1. Ax.

*12 Def. +Const,

Because the point F is the centre of the circle DKL, FD is equal to FK; but FD is equal to the Const. straight line A; therefore FK is equal to A. Again, because G is the centre of the circle LHK, GH is equal to GK; but GH is equal to C; therefore also GK is equal to C: and FG is equalt to B: therefore the three straight lines KF, FG, GK are respectively equal to the three A, B, C: and therefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines A, B, C, each to each. Which was to be done.

PROP. XXIII. PROB.

At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle; it is required to make an angle at the given point A in the given straight line AB, that shall be equal to the given rectilineal angle DCE.

In CD, CE, take any points D, E, and join DE; and make* the triangle AFG, the sides

*22. 1.

of which shall be equal to the three

straight lines CD, DE, EC, so that D CD be equal to AF, CE to AG, and

DE to FG: the angle FAG shall

be equal to the angle DCE.

C

H

Because DC, CE are equal to FA, AG, each to each, and the base DE to the base FG: the angle DCE is equal to the angle FAG. Therefore at the given ́ point A in the given straight line AB, the angle

*8. 1.

FAG is made equal to the given rectilineal angle DCE. Which was to be done.

After having made AF equal to CD, parts AH, FB of the given line must be taken equal respectively to CE, DE, in order to apply the construction in last proposition. If the given line be not sufficiently long for this, it must of course be produced.

PROP. XXIV. 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 one of them 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.

Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF: the base BC shall be greater than the base EF.

Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make* the angle EDG equal to

#23. 1. the angle BAC; and make DG equal to AC or *3.1. DF, join EG, GF, and let H be the point where EG is intersected either by DF or by DF prolonged.

+Hyp.

Because DE is not greater than DG the angle DGE cannot be greater than DEG; but DHG is greater than DEG, therefore DHG is greater than DGH, and, consequently, DG is greater than DH, but DF is equal to DG, therefore DF is greater than DH, so that the angle HGF is a part of DGF. Again, because AB is equal to Coust. DE, and AC† to DG, the two sides, BA, AC, are equal to the two ED, DG, each to each, and the angle BAC is equalt to the angle EDG; therefore the base BC is equal to the base EG. because DG is equal to DF, the angle DFG is equal* to the angle DGF; but it has been proved that DGF is

+Const.

#4. 1.

#5. 1.

And

19 Ax.

greater than the angle EGF; therefore the angle DFG is greater than EGF; therefore much more is the angle EFG greater than the angle EGF and because the angle EFG of the triangle EFG is greater than its angle EGF,

and that the greater angle is subtended by the #19. 1. greater side; therefore the side EG is greater than the side EF: but EG was proved to be equal to BC; therefore BC is greater than EF. Therefore if two triangles, &c.

Q. E. D.

PROP. XXV. THEOR.

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.

Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the base BC greater than the base EF: the angle BAC shall be greater than the angle EDF.

For, if it be not greater, it must A either be equal to it, or less than it: the angle BAC is not equal to the angle EDF, because then the base BC would be equal

*4. 1

+Hyp. to EF: but it ist not: B therefore the angle BAC

D

C E

is not equal to the angle EDF: neither is it less, because then the base BC would be less than the base EF: but it ist not; therefore the angle BAC is not less than the angle EDF: and it was shewn that it is

#24. 1. fHyp.