right angles: hence, AC is equal to AE; which was to be proved.

3°. It may be shown, as in the first case, that AD is equal to DF. Then, because the point C lies within the triangle ADF, the sum of the lines AD and DF is greater than the sum of the lines AC and CF (P. VIII.): hence, AD, the half of ADF, is greater than AC, the half of ACF; which was to be proved.

Cor. 1. The perpendicular is the shortest distance from a point to a line.

Cor. 2. From a given point to a given straight line, only two equal straight lines can be drawn; for, if there could be more, there would be at least two equal oblique lines on the same side of the perpendicular; which is impossible.

If a perpendicular is drawn to a given straight line at its middle point:

1o. Any point of the perpendicular is equally distant from the extremities of the line:

2°. Any point, without the perpendicular, is unequally distant from the extremities.

Let AB be a given straight line, C its middle point, and EF the perpendicular. Then any point of EF is equally distant from A and B; and any point without EF, is unequally distant from A and B.

1o. From any point of EF, as D, draw the lines DA and DB. Then DA and DB are equal (P. XV.): hence, D is equally distant from A and B; which was to be proved.

F

2o. From any point without EF, as I, IB. One of these lines, as IA, will cut point D; draw DB. Then, from what has just been shown, DA and DB are equal; but IB is less than the sum of ID and DB (P. VII.); and because the sum of ID and DB is equal to the sum of ID and DA, or IA, we have IB less than IA: hence, I is unequally distant from A and B; which was to be proved.

Cor. If a straight line, EF, has two of its points, E and F, each equally distant from A and B, it is perpendicular to the line AB at its middle point.

If two right-angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, each to each, the triangles are equal in all respects.

Let the right-angled triangles ABC and DEF have the hypothenuse AC equal to DF,

and the side AB equal to DE: A

then the triangles are equal

in all respects.

If the side BC is equal to

EF, the triangles are equal,

B

in accordance with Proposition X. Let us suppose then, that BC and EF are unequal, and that BC is the longer. On BC lay off BG equal to EF, and draw AG. The triangles ABG and DEF have AB equal to DE, by hypothesis, BG equal to EF, by construction, and the angles B and E

equal, because both are right angles; consequently, AG is equal to DF (P. V.). But, AC is equal to DF, by hypothesis: hence, AG and AC are equal, which is impossible (P. XV.). The hypothesis that BC and EF are unequal, is, therefore, absurd: hence, the triangles have all their sides equal, each to each, and are, consequently, equal in all respects; which was to be proved.

PROPOSITION XVIII. THEOREM.

If two straight lines are perpendicular to a third straight line, they are parallel.

Let the two lines AC, BD, be perpendicular to AB: then they are parallel.

B

D

For, if they could meet in a point O, there would be two perpendiculars OA, OB, drawn from the same point to the same straight line; which is impossible (P. XIV.): hence, the lines are parallel; which was to be proved.

DEFINITIONS.

E

If a straight line EF intersect two other straight lines AB and CD, it is called a secant, with respect to them. The eight angles formed about the points of intersection have different names, with respect to each other.

1o. INTERIOR ANGLES ON THE SAME

SIDE, are those that lie on the same

side of the secant and within the other two lines. Thus, BGH and GHD are interior angles on the same side.

2o. EXTERIOR ANGLES ON THE SAME SIDE are those that

lie on the same side of the secant and without the other two lines. Thus, EGB and DHF

4°. ALTERNATE EXTERIOR ANGLES are those that lie on opposite sides of the secant and without the other two lines. Thus, AGE and FHD are alternate exterior angles.

5°. OPPOSITE EXTERIOR AND INTERIOR ANGLES are those that lie on the same side of the secant, the one within and the other without the other two lines, but not adjacent. Thus, EGB and GHD are opposite exterior and interior angles.

If two straight lines meet a third straight line, making the sum of the interior angles on the same side equal to two right angles, the two lines are parallel.

Let the lines KC and HD meet the line BA, making the sum of the angles BAC and ABD equal to two right angles; then KC and HD are parallel.

Through G, the middle point

angles (P. I.); the sum of the angles FAG and GBD is equal to two right angles, by hypothesis: hence (A. 1),

GBE + GBD = FAG + GBD.

Taking away the common part GBD, we have the angle GBE equal to the angle FAG. Again, the angles BGE and AGF are equal, because they are vertical angles (P. II.): hence, the triangles GEB and GFA have two of their angles and the included side equal, each to each; they are, therefore, equal in all respects (P. VI.): hence, the angle GEB is equal to the angle GFA. But, GFA is a right angle, by construction; GEB must, therefore, be a right angle: hence, the lines KC and HD are perpendicular to EF, and are, therefore, parallel (P. XVIII.) ; which was to be proved.

Cor. 1. If two straight lines are cut by a third straight line, making the alternate angles equal to each other, th two straight lines are parallel.

E

Let the angle HGA be equal to GHD. Adding to both the angle HGB, we have,

A

B

H

HGA + HGB = GHD + HGB. But the first sum is equal to two right angles (P. I.): hence, the second sum is also equal to two right angles; therefore, from what has just been shown, AB and CD are parallel.

Cor. 2. If two straight lines are cut by a third, making the opposite exterior and interior angles equal, the two straight lines are parallel. Let the angles EGB and GHD be equal: Now EGB and AGH are equal, because they are vertical angles (P. II.); and consequently, AGH and GHD are equal: hence, from Cor. 1, AB and CD are parallel.