PROP. XIV. THEOR.

Two right lines (CB, BD), which, at the same point (B), and on different sides of a right line (AB), make with it, the adjacent angles (CBA, DBA), together equal to two right angles, are in the same right line.

If BD be not in the same right line with CB, let some other right line, as BE, be the production of CB; and, because the right line AB falls on the right line CBE, the angles CBA, ABE are equal C to two right angles (13. 1); but

B

the angles CBA, ABD are, by supposition, equal to two right angles; therefore the angles CBA, ABE together, and the angles CBA, ABD together, being each equal to two right angles, and all right angles being equal to each other (Theor. at 11. 1), are equal (Ax. 1); taking from each the common angle CBA, the remaining angles ABE, ABD are equal (Ax. 3), part and whole, which is absurd (Ax. 9): therefore BE is not the continuation of CB. In like manner it may be shewn, that no other right line, but BD, can be the continuation of CB; therefore BD is that continuation, and CB, BD are in the same right line.

PROP. XV. THEOR.

If two right lines (AB, CD) intersect each other, the vertical or opposite angles are equal; (AEC to BED, and AED to CEB,).

The angles AEC, AED, which AE makes with CD, are equal to two right angles [13. 1.] also the angles AED, DEB, which DE makes with AB, are equal to two right angles [13. 1]; therefore the angles AEC, AED together, are equal

to AED, DEB together (Theor. at 11. 1. and Ax. 1.): taking away the common angle AED, the remaining angles AEC and DEB are equal (Ax. 3). In like manner, the angles AED, CEB may be proved equal.

Cor. Two intersecting right lines (AB, CD), make angles, equal to four right angles.

Cor. 2-If more right lines be drawn to the point, wherein two right lines intersect each other, all the angles taken together, are equal to four right angles.

PROP. XVI. THEOR.

If any side (BC) of a triangle (ABC) be produced, the exterior angle (ACD) is greater than either of the interior remote angles (A or ABC)

Bisect AC in E (10. 1), join BE, on which produced take EF equal to BE (3. 1), and join CF.

In the triangles AEB, CEF, the sides CE, EF are severally equal to AE, EB [Constr.], and the angles

B

F

CEF, AEB, being vertical angles, are equal (15. 1); therefore the angles ECF and A are equal (4. 1); whence ACD, being greater than ECF [Ax. 9], is also greater than its equal A. In like manner, if AC be produced, as to G, the angle BCG may be proved to be greater than ABC; and therefore ACD, which is equal to BCG (15. 1), is also greater than ABC.

PROP. XVII. THEOR.

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

Produce any side BC, as to D, and the exterior angle ACD of the triangle ABC is greater than the interior remote angle B (16. 1); adding to each the angle ACB, the angles ACD, ACB together, are greater than the angles B and

A

B

•D

ACB together [Ax. 4]; but the angles ACD, ACB, which AC makes with BD, are together equal to two right angles (13. 1), therefore the angles B and ACB are together less than two right angles. In like manner it may be shown, that any other two angles of the triangle ABC are less than two right angles.

Cor.-If, in any triangle, one angle be obtuse or right, both the others are acute; and if two angles be equal, they are both

acute.

PROP. XVIII. THEOR.

If two sides (AB, AC), of a triangle (ABC), be unequal; the angle (ABC), opposite the greater side (AC), is greater than that (C), opposite the less (AB).

B

D

From the greater side AC, take away AD, equal and conterminous to the less AB [3. 1], and join BD.

Because the triangle ABD is isosceles, the angles ABD, ADB are equal [5. 1]; but the external angle ADB of the triangle BDC is greater than the internal remote angle C, [16. 1], therefore ABD is greater than C; of course ABC, which is greater than ABD [Ax. 9], is also greater than C.

PROP. XIX. PROB.

If two angles (B, C), of a triangle (ABC), be unequal; the side (AC) opposite the greater angle (B), is greater than that (AB), opposite the less (C).

A

B

If AC be not greater than AB, it is either equal to or less than it; it is not equal to AB, for then the angles ABC, ACB would be equal [5. 1], contrary to the supposition; it is not less than AB, for then the angle B would be less than the angle C [18. 1], which is also contrary to the supposition.

Cor. 1.-A perpendicular (CA), drawn from any point (C), to any right line (AB), is less than any other right line (CB), drawn from the same point, to the same right line.

For the angle CAB being right, CBA is acute [Cor. 17. 1], therefore CB is greater than CA (by this prop).

Cor. 2.—If perpendiculars let fall from two points (C, F), on two right lines [AB, DE], be equal; and from the same points, right lines [CB, FE], be drawn to points [B, E], in those right lines, at unequal distances [AB, DE], from the incidences [A. D] of the perpendiculars, that [FE], which is drawn to the most distant point [E], is greater than the other [CB].

On DE, take DG equal to AB [3. 1], and draw FG; in the triangles CAB, FDG, CA, AB and the angle CAB, are severally equal to FD, DG and the angle FDG, therefore FG is equal to CB [4. 1]; and the external angle FGE of the triangle FDG is greater than the interior FDG [16. 1]; whence, FDG being a right angle, FGE is obtuse, and therefore FEG acute [Cor. 17. 1], and so FE greater than FG [19. 1], or its equal CB.

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

C

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. 1]; 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