b

(3) Constr. EF and EC (3), and CF is common to the triangles DFC and EFC; therefore the angles DCF and ECF, opposite to the equal sides DF and EF, are equal (4), and therefore FC is perpendicular to the given right (5) Def. 11. line AB at the point Ĉ (5).

(4) Schol.

Fig. 25.
See N.

Schol. In the same manner a perpendicular can be erected at the extremity of a given line, by first producing the line.

To draw a perpendicular to a given indefinite right line (AB), from a point (C) given without it.

Take any point X on the other side of the given line, and from the centre C with the radius CX describe a circle, cutting the given line in E and F. Bisect EF in (1) Prop. 10. D (1), and draw from the given point to the point of bisection the right line CD: it is perpendicular to the given line.

(2) Def. 15.

(3) Constr.

(4) Schol.

Prop. 8.
(5) Def. 11.

Fig. 26.
See N.

(1) Def. 11.
(2) Prop. 11.

For draw CE and CF; and, in the triangles EDC and FDC, the sides EC and FC (2) are equal, and ED and FD (3) are equal, and CD is common; therefore the angles EDC and FDC, opposite to the equal sides EC and FC, are equal (4), and therefore DC is perpendicular to the given line AB (5).

PROP. XIII. THEOR.

When a right line (AB) standing upon another (DC) makes angles with it, they are either two right angles, or together equal to two right angles.

If the right line AB be perpendicular to DC, the angles ABC and ABD are right (1). If not, draw BE perpendicular to DC (2); and it is evident that the angles CBA and ABD together are equal to the angles CBE and EBD, and therefore equal to two right angles.

Cor. 1. If several right lines stand on the same right line at the same point, and make angles with it; all the angles taken together are equal to two right angles.

Cor. 2. Two right lines intersecting one another make angles, which, taken together, are equal to four right angles.

Cor. 3. If several right lines intersect one another in the same point, all the angles taken together are equal to four right angles.

PROP. XIV. THEOR.

If two right lines (CB and BD), meeting another right Fig. 27. line (AB) at the same point and at opposite sides, make angles with it, which are together equal to two right angles; those right lines (CB and BD) form one continued right line.

If it be possible, let BE and not BD be the continuation of the right line CB; then the angles CBA and ✰ ABE are equal to two right angles (1): but CBA and (1) Prop. 13. ABD are also equal to two right angles (2); therefore (2) Hypoth. CBA and ABD taken together are equal to CBA and ABE: take away from these equal quantities CBA, which is common to both, and ABE shall be equal to ABD, a part to the whole, which is absurd. Therefore BE is not the continuation of CB; and in the same manner it can be proved that no other line, except BD, is the continuation of it: therefore BD forms with BC one continued right line.

PROP. XV. THEOR.

If two right lines (AB and CD) intersect one another, Fig. 28. the vertical angles are equal (CEA to BED, and CEB a to AED).

Because the right line CE stands upon the right line AB, the angle AEC, together with the angle CEB, is equal to two right angles (1); and because (1)Prop. 13. the right line BE stands upon the right line CD, the angle CEB, together with the angle BED, is equal to two right angles (1); therefore AEC and CEB together are equal to CEB and BED: take away the common angle CEB, and the remaining angle AEC is equal to BED (2).

(2) Ax. 3.

Fig. 29.

(1) Prop. 10.

(2) Prop. 3.

(3) Constr.
(4) Prop.15.
(5) Prop. 4.

Fig. 30.

PROP. XVI. THEOR.

If one side (BC) of a triangle (BAC) be produced, the external angle (ACD) is greater than either of the internal opposite angles (A or B).

Bisect the side AC in E (1); draw BE, and produce it until EF be equal to BE (2), and join FC.

The triangles CEF and AEB have the sides CE and EF equal to the sides AE and EB (3), and the angle CEF equal to AEB (4); therefore the angles ECF and A are equal (5), and therefore ACD is greater than A. In like manner it can be shewn that, if AC be produced, the external angle BCG is greater than the angle B, and therefore that the angle ACD,_which is equal to BCG (4), is greater than the angle B.

Cor. 1. If, from any point B, two right lines be drawn to the same right line ED, one of them perpendicular to it, the other not; the perpendicular falls at the side of the acute angle.

For, if possible, let BA perpendicular to the line ED fall at the side of the obtuse angle BCE, then the (1) Def. 12. angle BAE is less than BCE (1); but BAE is greater (2) Prop.16. than the same angle BCE (2), which is absurd: BA therefore cannot fall at the side of the obtuse angle, and therefore must fall at the side of the acute angle.

Fig. 33.

(1) Ax. 11. (2) Prop.16.

Fig. 31.

Cor. 2. Two perpendiculars cannot be drawn from the same point B, to the same right line ED.

For, if possible, let the lines BA and BC be both perpendicular to ED, then the angle BAE is equal to the angle BCE (1); but it is also greater than it (2), which is absurd: therefore the right lines BA and BC cannot both be perpendicular to ED.

PROP. XVII. THEOR.

Any two angles of a triangle (BAC) are together less than two right angles.

Produce any side BC, then the angle ACD is greater (1) Prop. 16. than either the angle A or B (1); therefore ACB, toge

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 join BD. The triangle BAD being isosceles (2), the angles ABD and ADB are equal (3); but ADB is greater than the internal angle ACB (4); therefore ABD is 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.

(1) Prop. 3.

(2) Constr.
(3) Prop. 5.
(4) Prop. 16.

If in any triangle (BAC) one angle (ABC) be greater Fig. 32. 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 to, or less or greater than AB.

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

It is not less than AB, because the angle ABC would

then be less than ACB (2); which is contrary to the (2) Prop. 18. hypothesis.

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

Cor. 1. 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 the less angle, and therefore is less than BA, which is opposite to the greater (2).

Cor. 2. If a line AD be drawn from the vertex of any triangle BAD, to the base BC, it will be less than the greater of the sides containing the angle BAC, if those sides be unequal; or than either of them, if they be equal.

Let the side AB be greater than AC or equal to it, then the angle ACB will be greater than or equal to ABC (1); but the angle ADB is greater than ACD (2), and therefore greater than ABD; therefore the side AB is greater than AD (3). If AC be equal to AB, it is evident it will also be greater than AD.

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 DAC (2); but BÃD is equal to DAC (3), therefore BDA is greater than BAD, and therefore the side BA is greater than BD (4): in the same manner it can be demonstrated that the side AC is greater than CD; therefore the two sides BA and AC, taken together, are greater than BD, DC, or the third side BC. Thus by bisecting any angle, it can be demonstrated that the sides containing that angle are greater than the third side.

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