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 HGA 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 IIGL,

E

L

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

C

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.

Let AB and CD be respectively parallel to EF: then will they be parallel to each other.

For, draw PR perpendicular to EF; then will it be perpendicular to AB, and also to CD (P. XX., C. 1):

hence, AB and CD are perpendicu

E

R

F

C

1Q

D

A

P

B

lar to the same straight line, and consequently, they are par allel 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 every

where equally distant.

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. 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,

Draw also VG.

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 in- L terior angle ALE (P. XX., C. 3); and because BC and LF are parallel, the exterior angle ALE is equal to its op

B

D

-F

E

posite interior angle ABC : hence, DEF is equal to ABC; which was to be proved.

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

A

G H

M

B

K

Prolong GH to M. Then, because KH and BM are parallel, the exterior angle GIIK is equal to its opposite interior angle HMB ; and because IIM and BC are parallel, the angle HMB is equal to its alternate angle MBC (P. XX., C. 2): hence, GHK is equal to ABC; which was to be proved.

Cor. The opposite angles of a parallelogram are equal.

PROPOSITION XXV. THEOREM.

In any triangle, the sum of the three angles is equal to two right angles.

Let CBA be any triangle: then will the sum of the angles C, A, and B, be equal to

two right angles.

For, prolong CA to D, and draw AE parallel to BC.

E

D

In like manner,

Then, since AE and CB are parallel, and CD cuts them, the ex terior angle DAE is equal to its opposite interior angle C (P. XX., C. 3). since AE and CB are parallel, and AB cuts them, the alternate angles ABC and BAE are equal: hence, the sum of the three angles of the triangle BAC, is equal to the sum of the angles CAB, BAE, EAD; but this sum is equal to two right angles (P. I., C. 2); consequently, the sum of the three angles of the triangle, is equal to two right angles (A. 1); which was to be proved.

Cor. 1. Two angles of a triangle being given, the third will be found by subtracting their sum from two right angles.

Cor. 2. If two angles of one triangle are respectively equal to two angles of another, the two triangles are mutually equiangular.

Cor. 3. In any triangle, there can be but one right angle; for if there were two, the third angle would be zero. Nor can a triangle have more than one obtuse angle.

Cor. 4. In any right-angled triangle, the sum of the acute angles is equal to a right angle.

Cor. 5. Since every equilateral triangle is also equiangular (P. XI., C. 1), each of its angles will be equal to the third part of two right angles; so that, if the right angle is expressed by 1, each angle, of an equilateral triangle, will be expressed by 2.

Cor. 6. In any triangle ABC, the exterior angle BAD ís equal to the sum of the interior opposite angles B and C. For, AE being parallel to BC, the part BAE is equal to the angle B, and the other part DAE, is equal to the angle C.

PROPOSITION XXVI. THEOREM.

The sum of the interior angles of a polygon is equal to two right angles taken as many times as the polygon has sides, less two.

Let ABCDE be any polygon: tnen will the sum of its interior angles A, B, C, D, and E, be equal to two right angles taken as many times as the polygon has sides, less

two.

E

D

From the vertex of any angle A, draw diagonals AC, AD. The polygon will be divided into as many triangles, less two, as it has sides, having the point 4 for a common vertex, and for bases, the sides of the polygon, except the two which form the angle A. It is evident, also, that the sum of the angles of these triangles does not differ from the sum of the angles of the polygon: hence, the sum of the angles of the polygon is equal to two right angles, taken as many times as there are triangles; that is, as many times as the polygon has sides, less two; which was to be proved.