G (10. 1), and join CG, which is the perpendicular required. For, CE, CF being drawn, in the triangles CGE, CGF, the sides GE, GF are equal, by construction, CG common, and the bases CE, CF equal, being radiuses of the same circle EDF (Def. 10); therefore the angles CGE, CGF are equal (8. 1), and of course CG is perpendicular to AB (Def. 20).

give

us the pleural of Causa", (it must be causa

PROP. XIII. THEOR.

The angles (ABC, ABD), which one right line (AB), makes with another (CD), on one side of it, are together equal to two right angles.

If the angles ABC, ABD be equal, (see figure 1), they are both right angles (Def. 20), and therefore together equal to two right angles; if unequal, (see fig. 2), draw BE perpendicular to CD (11. 1), and EBC, EBD are both right angles (Constr.), and therefore together equal to two right angles; but, because the angle ABC is equal to all its parts, the two angles ABE, EBC taken together (Cor. Ax.8), if to each be added the angle ABD, the two angles ABC, ABD are together equal to the three angles CBE, EBA, ABD (Ax. 2); again, because the whole angle EBD is equal to all its parts EBA, ABD (Cor. Ax. 8), if to each CBE be added, the two angles CBE, EBD, are together equal to the three angles CBE, EBA, ABD, (Ax. 2): whence, the two angles CBA, ABD, and the two angles CBE, EBD being each equal to the three angles CBE, ESA, ABD, are equal to each other (Ax. 1): but CBE, EBD are right angles (Constr.), and therefore together equal to two right angles, therefore CBA, ABD are also together equal to two right angles.

Cor.--If more right lines stand on the same right line (CD), on the same side of it, and at the same point (B), not being an extreme; they make angles equal to two right angles,

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 BĎ 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. 1-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 B AE, EB [Constr.], and the angles

G

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

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.

1

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).

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.