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

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

angles.

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

Then, since AE and CB are parallel, and CD cuts them, the exterior angle DAE is equal to its opposite

E

interior angle C (P. XX., C. 3). In like manner, 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 may 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 is 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 is expressed by .

Cor. 6. In any triangle ABC, the exterior angle BAD is 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.

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

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

E

B

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 A 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, less two, as the polygon has sides; which was to be proved.

Cor. 1. The sum of the interior angles of a quadrilateral is equal to two right angles taken twice; that is, to four right angles. If the angles of a quadrilateral are equal, each is a right angle.

Cor. 2. The sum of the interior angles of a pentagon is equal to two right angles taken three times; that is, to six right angles: hence, when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or to of one right angle.

Cor. 3. The sum of the interior angles of a hexagon is equal to eight right angles: hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or of one right angle.

Cor. 4. In any equiangular polygon, any interior angle is equal to twice as many right angles as the figure has sides, less four right angles, divided by the number of angles.

The sum of the exterior angles of a polygon is equal to four right angles.

Let the sides of the polygon ABCDE be prolonged, in the same order, forming the exterior angles a, b, c, d, e; then the sum of these exterior angles is equal to four right angles.

For, each interior angle, together with the corresponding exterior angle, is equal

B

to two right angles (P. I.); hence, the sum of all the interior and exterior angles is equal to two right angles taker.

But the sum of

as many times as the polygon has sides. the interior angles is equal to two right angles taken as many times, less two, as the polygon has sides: hence, the sum of the exterior angles is equal to two right angles taken twice; that is, equal to four right angles; which was to be proved.

In any parallelogram, the opposite sides are equal, each to

each.

Let ABCD be a parallelogram: then AB is equal to DC, and AD to BC.

parallel, the

its alternate

D

B

For, draw the diagonal BD. Then, because AB and DC are angle DBA is equal to angle BDC (P. XX., C. 2); parallel, the angle BDA is DBC. The triangles ABD and CDB, have, therefore, the angle DBA equal to CDB, the angle BDA equal to DBC, and the included side DB common; consequently, they are equal in all respects: hence, AB is equal to DC, and AD to BC; which was to be proved.

and, because AD and BC are equal to its alternate angle

Cor. 1. A diagonal of a parallelogram divides it into two triangles equal in all respects.

Cor. 2. Two parallels included between two other parallels, are equal.

Cor. 3. If two parallelograms have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, they are equal.

PROPOSITION XXIX. THEOREM.

If the opposite sides of a quadrilateral are equal, each to each, the figure is a parallelogram.

In the quadrilateral ABCD, let AB be equal to DC, and AD to BC: then is it a parallelogram.

D

A

B

Draw the diagonal DB. Then, the triangles ADB and CBD, have the sides of the one equal to the sides of the other, each to each; and therefore, the triangles are equal in all respects: hence, the angle ABD is equal to the angle CDB (P. X., S.); and consequently, AB is parallel to DC (P. XIX., C. 1). The angle DBC is also equal to the angle BDA, and consequently, BC is parallel to AD: hence, the opposite sides are parallel, two and two; that is, the figure is a parallelogram (D. 28); which was to be proved.

If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram.

In the quadrilateral ABCD, let AB be equal and parallel to DC: then the figure is a parallelogram.

D

B

Draw the diagonal DB. Then, because AB and DC are parallel, the angle ABD is equal to its alternate angle CDB. Now, the triangles ABD and CDB have the side DC equal to AB, by hypothesis, the side DB common, and the included angle ABD equal to BDC, from what has just been shown;