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

GBE + GBD = FAG + GBD.

LS

Taking from both 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 their parts (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 both perpendicular to EF, and are, therefore, parallel (P. XVIII.); which was to be proved.

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

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

PROPOSITION XX. THEOREM.

I a straight line intersect two parallel straight lines, the sum of the interior angles on the same side will be equal to two right angles.

Let the parallels AB, CD, be cut by the secant line FE: then will the sum of HGB and GHD be equal to

two right angles.

A

C

E

I

G

-B

L

-D

H

F

For, if the sum of HGB and GHD is not equal to two right angles, let IGL be drawn, making the sum of HGL and GHD equal to two right angles; then IL and CD will be parallel (P. XIX.); and consequently, we shall have two lines GB, GL, drawn through the same point G and parallel to CD, which is impossible (A. 13): hence, the sum of HGB and GHD, is equal to two right angles; which was to be proved.

In like manner, it may be proved that the sum of HGA and GHC, is equal to two right angles.

Cor. 1. If HGB is a right angle, GHD will be a right angle also hence, if a line is perpendicular to one of two parallels, it is perpendicular to the other also.

Cor. 2. If a straight line meet two parallels, the alternate angles will be equal.

are equal. Taking away the common part BGH, there remains the angle GHD equal to HGA. In like manner, it may be shown that BGH and GHC are equal.

Cor. 3. If a straight line meet two parallels, the opposite exterior and interior angles will be equal. The angles DHG and IIGA are equal, from what has just been shown. The angles HGA and BGE are equal, because they are vertical hence, DHG and BGE are equal. In like manner,

it may be shown that CHG and AGE are equal.

Scholium. Of the eight angles formed by a line cutting two parallel lines obliquely, the four acute angles are equal, and so, also, are the four obtuse angles.

PROPOSITION XXI. THEOREM.

If two straight lines intersect a third line, making the sum of the interior angles on the same side less than two right angles, the two lines will meet if sufficiently produced.

Let the two lines CD, IL, meet the line EF, making the sum of the interior angles HGL, GHD, less than two right angles: then will IL and CD meet if sufficiently produced.

For, if they do not meet, they must be parallel (D. 16). But, if they were parallel, the sum of the interior angles HGL,

GHD, would be equal to two right angles (P. XX.), which is

C

H

E

G

-L

-D

contrary to the hypothesis: hence,

IL, CD, will meet if sufficiently produced; which was to be proved.

Cor. It is evident that IL and CD, will meet on that side of EF, on which the sum of the two angles is less than two right angles.

PROPOSITION XXII. THEOREM.

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

lar to the same straight line, and consequently, they are parallel to each other (P. XVIII.); which was to be proved.

PROPOSITION XXIII. THEOREM.

Two parallels are everywhere equally distant.

Let AB and CD be parallel: then will they be everywhere equally distant.

CH

A

F

GD

B

From any two points of AB, as F and E, draw FH and EG perpendicular to CD; they will also be perpendicular to AB (P. XX., C. 1), and will measure the distance between AB and CD, at the points F and E. Draw also FG. The lines FH and EG are parallel (P. XVIII.): hence, the alternate angles HFG and FGE are equal (P. XX., C. 2). The lines AB and CD are parallel, by hypothesis: hence,

the alternate angles EFG and FGH are equal. The triangles FGE and FGH have, therefore, the angle HGF equal to GFE, GFH equal to FGE, and the side FG common; they are, therefore, equal in all their parts (P. VI.) : hence, FH is equal to EG; and consequently, AB and CD are everywhere equally distant; which was to be proved.

PROPOSITION XXIV. THEOREM.

If two angles have their sides parallel, and lying either in the same, or in opposite directions, they will be equal.

1o. Let the angles ABC and DEF have their sides parallel, and lying in the same direction: then will they be equal.

Prolong FE to L. Then, because DE and AL are parallel, the exterior angle DEF is equal to its opposite interior angle ALE (P. XX., C. 3); and because BC and LF are parallel, the exterior angle ALE is equal to its op

A

L

-F

E

B

C

posite interior angle ABC: hence, DEF is equal to

ABC; which was to be proved.

2o. Let the angles ABC and GHK have their sides parallel, and lying in opposite directions: then will they be equal.

Prolong GH to M. Then, because KH and BM are parallel, the exterior angle GHK is equal to its and because HM and BC is equal to its alternate angle GHK is equal to ABC;

A

G H

M

B

C

K

opposite interior angle HMB; are parallel, the angle HMB MBC (P. XX., C. 2): hence, which was to be proved.

Cor. The opposite angles of a parallelogram are equal.