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ther with either A or B, is less than the same angle ACB, together with ACD, that is, less than two right angles (2). In the same manner, if CB be produced (2) Prop. 13. from the point B, it can be demonstrated that the angle ABC together with the angle A, is less than two right angles: therefore any two angles of the triangle are less than two right angles.

Cor. If any angle of a triangle be obtuse or right, the other two angles are acute: and if two angles be equal to one another, they are acute.

PROP. XVIII. THEOR.

In any triangle (BAC) if one side (AC) be greater Fig. 32. than another (AB) the angle opposite to the greater side

is greater than the angle which is opposite to the less.

From the greater side AC cut off the part AD

equal to the less (1), and conterminous with it, and (1) Prop. 3. join BD.

The triangle BAD being isosceles (2), the angles (2) Constr. ABD and ADB are equal (3), but ADB is greater (3) Prop. 5. than the internal angle ACB (4), therefore ABD is (4) Prop. 16. greater than ACB, and therefore ABC is greater than ACB: but ABC is opposite the greater side AC, and ACB is opposite the less AB.

PROP. XIX. THEOR.

If in any triangle (BAC), one angle (ABC) be Fig. 32. greater than another (C), the side (AC) which is opposite to the greater angle is greater than the side (AB), which is opposite to the less.

For the side AC is either equal or less or greater than AB.

32

B

It is not equal to AB because the angle ABC would then be equal to ACB (1), which is contrary (1) Prop. 5. to the hypothesis.

It is not less than AB, because the angle ABC would then be less than ACB (2) which is contrary (2) prop. 18. to the hypothesis,

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Fig. 33.

(1) Cor.

Since, therefore, the side AC is neither equal to, nor less than AB, it is greater than it.

Cor. If from the same point B two right lines BC and BA be drawn to the same right line ED, of which one BC is perpendicular to ED, the other not, BC is less than BA.

For in the triangle ABC the angle BAC is acute, since BCA is right (1), therefore BC is opposite to Prop. 17. the less angle, and therefore is less than BA, which (2) Prop.19. is opposite to the greater (2).

Fig. 34.
See N.

(1) Prop.9.

PROP. XX. THEOR.

Any two sides (BA and AC) of a triangle (BAC) are together greater than the third side (BC).

Bisect the angle BAC by the right line AD (1); the external angle BDA is greater than the internal (2) Prop.16. DAC (2), but BAD is equal to DAC (3), therefore (3) Constr. BDA is greater than BAD, and therefore the side (4) Prop. 19 BA is greater than BD (4): in the same manner the side AC can be demonstrated to be greater than CD ; therefore the two sides BA and AC taken together are greater than BD, DC or the third side BC. Thus bisecting any angle, it can be demonstrated that the sides containing that angle are greater than the third side.

Fig.35.

(1) Prop.20

A

Schol. Hence it is evident, that the difference between two sides of any triangle, is less than the third side.

PROP. XXI. THEOR.

Two right lines (DB and DC) drawn to a point (D) within a triangle (BAC) from the extremities of any side (BC), are together less than the sam of the two other sides of the triangle (AB and C), but contain a greater angle.

Produce BD to E. The sides BA and AE of the triangle ABE are greater than the third side BE(1), add EC to each, and the sides BA and AC are greater than

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BE and EC; but the sides DE and EC of the triangle EDC are greater than the third side DC (1), add BD to each, and BE and EC are greater than BD and DC, but BA and AC are greater than BE and EC, therefore BA and AC are greater than BD and DC.

B

15

Because the external angle BDC is greater than the internal DEC (2), and for the same reason DEC (2) Prop. 18. is greater than BAE, the angle BDC is greater than the angle BAE.

PROP. XXII. PROB.

Given three right lines (A, B and C) of which any two together are greater than the third, to construct a Fig. 36. triangle whose sides shall be respectively equal to the See N. given lines.

From any point D draw the right line DE equal to one of the given lines A (1), and from the same point draw DG equal to another of the given lines B (1), and from the point E draw EF equal to C (1). (1) Prop. 2. From the centre D with the radius DG describe a circle, from the centre E with the radius EF describe another circle (2), and from the point of intersec- (2) Post. 3. tion K draw KD and KE. It is evident that the sides DE, DK and KE of the triangle DKE are equal to the given right lines A, B and C.

Cor. In this manner a triangle can be constructed equal to a given one, by constructing a triangle whose sides shall be equal to the sides of the given, for this triangle is equal to the given one (3).

PROP. XXIII. PROB.

(3) Schol. Prop. 8.

At a given point (B), in a given right line (BE), to Fig. 37. make an angle equal to a given angle (C).

In the sides of the given angle take any points D and F, join DF and construct a triangle EBA which shall be equilateral with the triangle DCF, and whose sides AB and EB meeting at the given point B

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B

39

(1) Prop. 22 shall be equal to FC and DC of the given angle C (1). The angle EBA is equal to the given angle DCF.

For as the triangles DCF and EBA have all their sides respectively equal, the angles FCD and ABE (2) Prop. 8. opposite the equal sides DF and EA are equal (2).

Fig.38
See N.

38

PROP. XXIV. THEOR.

If two triangles (EFD, BAC) have two sides of the one respectively equal to two sides of the other, (FE to AB, and FD to AC), and if one of the angles (BAC) contained by the equal sides be greater than the other (EFD), the side (BC) opposite to that greater angle is greater than the side (ED), which is opposite to the less angle.

From the point A draw the right line AG, making with the side AB, which is not the greater, an angle (1) Prop. 23. BAG equal to the angle EFD (1), make AG equal (2) Prop. 3. to FD (2), and join BG, GC.

(3) Constr.
& hypoth.

Since the right lines AG and AC are equal (3), the angles ACG and AGC are equal (4); but the angle (4) Prop.5. BGC is greater than AGC, therefore greater than ACG, and therefore greater than BCG. Then in the triangle BGC, the angle BGC is greater than (5) Prop. 19. BCG, therefore the side BC is greater than BG (5); (8) Constr. but BG is equal to ED (6), and therefore BC is & Prop. 4. greater than ED.

Fig. 39.

AA

PROP. XXV. THEOR.

If two triangles (BAC and EFD) have two sides of the one respectively equal to two sides of the other (BA to EF, and AC to FD), and if the third side of the one (BC) be greater than the third side (ED) of the other, the angle (A) opposite to the greater side is greater than the angle (F), which is opposite to the less.

The angle A is either equal to the angle F, or less than it, or greater than it.

It is not equal, for if it were, the side BC would be

equal to the side ED (1), which is contrary to the (1) Prop. 4. hypothesis.

It is not less, for if it were, the side BC would be

less than the side ED (2), which is contrary to the (2) Prop. 24. hypothesis.

Since, therefore, the angle A is neither equal to, nor less than F, it is greater.

PROP. XXVI. THEOR.

If two triangles (BAC DEF) have two angles of Fig. 40. the one respectively equal to two angles of the other See N. (B to D, and C to F), and a side of the one equal to a side of the other, whether it be adjacent (BC to DF) or opposite (BA to DE) to those equal angles, the remaining sides and angles are respectively equal to one another.

First let the equal sides be BC and DF which are adjacent to the equal angles, then the side BA is equal to the side DE.

re

& hypoth.

For, if it be possible, let one of them BA be greater than the other, make BG equal to DE, and join In the triangles GBC, EDF the sides G respectively equal to the sides ED, DF (the) Constr. angle Bis equal to the angle D (2), therefore the angles (2) hypoth BCG and DFE are equal (3), but the angle BCA is (3) prop. 4. also equal to DFE (4), therefore the angle BCG is (4) hypoth. equal to BCA (5), which is absurd: neither of the (5) ax. 1. sides BA and DE, therefore, is greater than the other, therefore they are equal; and also BC and DF are equal (6), and the angles B and D (6), therefore the (6) hypoth. side AC is equal to the side EF, as also the angle A to (7) prop. 4. the angle E (7).

Next, let the equal sides be BA and DE, which are opposite to the equal angles C and F, and the sides BC and DF shall also be equal.

For, if it be possible, let one of them BC be greater than the other, make BH equal to DF, and join AH. In the triangles ABH, EDF, the sides AB, BH are (8) Constr. respectively equal to the sides ED,DF,(8), and the angle & hypoth B is equal to the angle D (9), therefore the angles AHB (9) hypoth

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