PROP. XX. THEOR.

Any two sides (as BA, AC) of a triangle (ABC), are greater than the remaining side (BC).

On either of the sides to be proved greater than the remaining one, as BA, produced, take AD equal to the other AC, and join CD.

B

A

The angles D and ACD at the base of the isosceles triangle ACD are equal [5. 1], whence BCD, being greater than ACD [Ax. 9], is greater than the angle D; therefore, in the triangle BCD, the side BD, opposite the greater angle BCD, is greater than the side BC, opposite the less angle D [19. 17; but, because AD is equal to AC, BD is equal to BA and AC together [Ax. 2], therefore BA and AC together are greater than BC.

PROP. XXI. THEOR.

Two right lines (DB, DC), drawn from any point (D), within a triangle (ABC), to the extremes of any side (BC), are together less than the other sides (AB, AC) of the triangle, but contain a greater angle (BDC).

E

Produce BD to meet AC in E, and because the sides BA, AE of the triangle BAE, are greater than the remaining side BE (20. 1), adding to each EC, the right lines BA and AC, are greater than BE and EC [Ax. 4]; but the sides DE, EC of the triangle DEC, are greater than the remaining side DC (20. 1), therefore, adding to each BD, the right lines BE and EC are greater than BD and DC [Ax. 4]; but it has been proved, that BA and AC ́are greater than BE and EC, therefore BA and AC are also greater than BD and DC.

B

And, because the external angle BDC, of the triangle DEC, is

greater than the internal remote angle BEC (16. 1), and the external angle BEC, of the triangle ABE, is greater than the internal remote angle A, the angle BDC is greater than A.

PROP. XXII. PROB.

Three right lines (A, B and C) being given, any two of which are greater than the remaining one, to make a triangle having its sides severally equal to these right lines.

At any point D, put the right line DF equal to A (2. 1), on which produced take FG equal to B, and GH to C (3. 1); from the centre F, at the distance FD, describe the circle DKM, and from the centre G, at the distance GH, describe the circle NKH [Post. 3]; and, since A and C, or their equals FM, NG together, are greater than B or FG [Hyp.], taking NG from each, FM is greater than FN [Ax. 5]; and B and C, or their equals FG, GH together, or FH, is greater than A or FM [Hyp.]; whence, FM being greater than FN and less than FH, the point M is between the points N and H in like manner, the point N may be shewn to be between the points D and M: whence, the circles DKM, NKH, being continued lines between the points D and M, and the points N and H, on each side of the right line DE [Def. 10], intersect each other on each side thereof, as in K and L : join FK, KG; the triangle FKG has its sides severally equal to the given right lines A, B and C.

For, because F is the centre of the circle DKM, FK is equal to FD [Def. 10], or its equal [Constr.] A; in like manner, GK may be proved equal to C; and FG is equal to B (Constr.)

Schol. The reasoning in this proposition to prove the intersection of the circles, is accommodated to the figure, but the same reasoning is manifestly applicable to all possible cases. (See note on this proposition.)

PROP. XXIII. PROB.

To a given right line (AB), at a given point therein (A), to make a rectilineal angle, equal to a given rectilineal angle (C).

Take any points, D, E in the legs of the given angle C, join DE, and let the triangle FAG be made equilateral to the triangle DCE, so that its sides FA, AG meeting at the given point A may be severally equal to the legs DC, CE of the given angle C, and one of these sides AF may be taken on the given right line AB (22. 1). The angle A is equal to the given angle C.

For because the triangles FAG, DCE are mutually equilateral, the angles A, C, opposite the equal sides FG, DE, are equal (8. 1).

PROP. XXIV. THEOR.

If two triangles (ABC, DEF), have two sides (AB, AC) of the ore, severally equal to two sides (DE, DF) of the other, but the angles (A and EDF) contained by these equal sides unequal, (the angle A being greater than the angle EDF,) the base (BC) of that triangle (ABC) which has the greater angle (A), is greater than the base (EF) of the other.

At the point D, with the right line DE, make the angle EDG equal to the angle A [23. 1], take DC equal to AC, and draw EG.

Because the angle EDG or A is greater than EDF, the right line DG falls without the triangle EDF, and the point F falls either below the right line EG, or on it, or above it.

If the point F fall below EG, draw FG; and because DF, DG are equal, the angles DFG, DGF are also equal [5. 1]; whence the angle EFG being greater than DFG, and DGF than EGF (Ax. 9), the angle EFG is greater than EGF, therefore EG is greater than EF [19. 1.]; but, because ED, DG and the included angle EDG are severally equal to BA, AC and the included angle A, EG is equal to BC [4. 1]; therefore BC is also greater than EF.

If the point F fall on EG, the right line EG, or its equal BC, is, in that case, greater than EF [Ax. 9].

If F falls above EG, the right lines DG, GE together are greater than DF, FE together [21. 1]; taking from each the equals DG, DF, the residue EG, or its equal BC, is greater than the residue EF.

PROP. XXV. THEOR.

If two triangles (ABC, DEF), have two sides (AB, AC) of the one, severally equal to two sides (DE, DF) of the other, but the bases or third sides (BC, EF) unequal (BC being greater than EF); the angle (A) opposite the greater base (BC), s greater than that (D) opposite the less (EF).

For if the angle A be not greater than D, iti s either equal to or less than it.

The angle A is not equal to D, for, if it were, the sides BA, AC being severally equal to ED, DF, the bases BC, EF would be equal [4. 1]; contrary to the hypothesis.

The angle A is not less than D, for, if it were, the sides BA, AC being severally equal to ED, DF, the base BC would be less than the base EF [24. 1]; which is again contrary the bypothesis.

Therefore the angle A, being neither equal to nor less than D, is greater than it.

PROP. XXVI. THEOR.

If two triangles (ABC, DEF), have two angles (A and ABC) of one, severally equal to two angles (D and E) of the other, and one side of one to one side of the other, namely either the sides (AB, DE) between the equal angles, or those (AC, DF) opposite to equal angles; the remaining sides are severally equal to each other, and the remaining angle (ACB) of the one, to the remaining angle (F) of the other.

Let first, AB be supposed equal to DE, then is AC equal to DF.

A

For if not, let one of hem, as AC,be, if possible, the greater, and on AC take AG equal A to DF, and draw BG.

D

E

In the triangles ABG, DEF, the sides AB, AG and the angle A, are severally equal to the sides DE, DF and the angle D [Constr. and Hyp.]; therefore the angles ABG and E are equal (4. 1); but the angles ABC and E are equal [Hyp.]; therefore the angles ABG, ABC, being each equal to E, are equal to each other [Ax. 1], part and whole, which is absurd [Ax. 9]; therefore the sides AC, DF are not unequal, they are therefore equal; whence AB being equal to DE, and the angle A to the angle D [Hyp.], BC is equal to EF, and the angle ACB to the angle F (4. 1).

Let now AC, DF, opposite the equal angles, ABC and E, be supposed equal, then is AB equal to DE.

For if not, let one of them, as AB, be the greater, and on AB take AH equal to DE, and draw CH.

In the triangles AHC, DEF, the sides AH, AC and the angle A, are severally equal to the sides DE DF and the angle D[Constr. and Hyp.], therefore the angles AHC and E are equal [4. 1]; but the angles ABC and E are equal [Hyp.], therefore the angles AHC, ABC, being each equal to E, are equal to each other [Ax. 1], namely, the external angle AHC of the triangle CHB to the internal remote angle ABC,