DEH-AGE+DEH. But AGE+DEH is equal to two right-angles; (Prop. 23;) hence DEH+BAC is equal to two right-angles.

PROPOSITION XXVIII. THEOREM.

In every triangle BAC, the sum of the three angles is equal to two right-angles.

Produce the side CA towards D; and at the point A, draw AE parallel to BC.

Since AE, CB are parallel, and CAD cuts them, the exterior angle

B

E

N

D

A

DAE will be equal to its interior opposite one ACB; (Prop. 24;) in like manner, since AE, CB are parallel, and AB cuts them, the alternate interior angles ABC, BAE will be equal: hence ACB+ABC-DAE+BAE. Add BAC to each, and ACB+ABC+BAC=DAE+BAE+BAC. But DAE+BAE+BAC is equal to two right-angles; (Prop. 1. Cor. 3;) therefore the sum of the three angles ACB, ABC, BAC, is equal to two right-angles. Hence,

The sum of the three angles of every triangle is equal to two right-angles.

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

Cor. 2. If two angles of one triangle are respectively equal to two angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular.

Cor. 3. In any triangle there can be but one right-angle ; for if there were two, the third angle must be nothing. Still less can a triangle have more than one obtuse angle.

Cor. 4. In every right-angled triangle, the sum of the two acute angles is equal to one right-angle.

Cor. 5. Since every equilateral triangle, is also equiangular, (Prop. 11. Cor.,) each of its angles will be equal to the third part of two right-angles; so that if the right-angle is expressed by unity, the angle of an equilateral triangle will be expressed by .

Cor. 6. In every triangle ABC, the exterior angle BAD is equal to the sum of the two 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 all the interior angles of a polygon ABCDEFG, is equal to as many times two right-angles, as the polygon has sides less two.

B

C

D

G F

E

For, by drawing diagonals AC, AD, AE, AF, from the vertex of any angle A to each of the opposite angles, it is evident that the A polygon will be divided into as many triangles as it has sides less two; and that the sum of the angles of all the triangles, is the same as the sum of all the angles of the polygon. But the sum of the angles of any triangle is equal to two right-angles; hence the sum of the angles of all the triangles is equal to as many times two right-angles as there are triangles; i. e. as the polygon has sides less two. Therefore,

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

two.

Cor. 1. The sum of the angles in a quadrilateral is equal to two right-angles multiplied by 4-2, i. e. to four rightangles hence if all the angles of a quadrilateral are equal, each of them will be a right-angle. (Def. 18.)

Cor. 2. The sum of the angles of a pentagon is equal to two right-angles multiplied by 5-2, i. e. 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 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.

PROPOSITION XXX. THEOREM.

The opposite sides and angles of a parallelogram are equal.

A

D

B

Draw the diagonal DB. The triangles ADB, DBC have a common side BD; since AD, BC are parallel, they have also the angle ADB=DBC; (Prop. 24;) and since AB, CD are parallel, the angle, ABD=BDC: hence these triangles are equal; (Prop. 6;) consequently the side AB, opposite the angle ADB, is equal to the side DC, opposite the angle DBC; and the sides AD, BC are also equal; therefore the opposite sides of a parallelogram are equal.

Again, since the triangles are equal, it follows that the angle A is equal to the angle C; and also that the angle ADC, composed of the two ADB, BDC, is equal to ABC, composed of the two DBC, ABD: therefore the opposite angles of a parallelogram are equal. Hence, The opposite sides, &c. Cor. 1. The diagonal of a parallelogram bisects it.

Cor. 2. Also, two parallels AB, CD, included between two other parallels AD, BC, are equal.

If the opposite sides of a quadrilateral ABCD, are equal, viz. AB to CD, and AD to BC, the equal sides will be parallel, and the figure will be a parallelogram.

For, having drawn the diagonal BD, the triangles ABD, BDC have all the sides of the one equal to the corresponding sides of the other; therefore they are equal; consequent

A

D

B

ly the angle ADB, opposite the side AB, is equal to DBC, opposite CD; (Prop. 10. Sch.;) hence the side AD is parallel to BC. (Prop. 21.) For a like reason, AB is parallel to CD: therefore the quadrilateral ABCD, is a parallelogram. Hence,

If the opposite sides of any quadrilateral are equal, the equal sides will be parallel, and the figure will be a parallelogram.

If two opposite sides AB, CD of a quadrilateral, are equal and parallel, the remaining sides will also be equal and parallel, and the figure ABCD will be a parallelogram.

D

Draw the diagonal BD. Since AB is parallel to CD, the alternate angles ABD, BDC are equal; (Prop. 24;) moreover, the side BD is common, and the side AB=CD: hence the triangle ABD is equal to DBC; (Prop. 5;)

A

B

therefore the side AD=BC, the angle ADB DBC, and consequently AD is parallel to BC. But by hypothesis the side AB is equal and parallel to CD; therefore the figure ABCD is a parallelogram. Hence,

If two opposite sides of any quadrilateral are equal and parallel, the remaining sides will be equal and parallel, and the figure will be a parallelogram.

PROPOSITION XXXIII.

THEOREM.

The two diagonals AC, DB of a parallelogram, bisect each

other.

For, comparing the triangles ADE, CEB, we find the side AD-CB, the D angle ADE CBE, (Prop. 24,) and the angle DAE ECB; hence those

triangles are equal; (Prop. 6;) hence

A

C

B

AE, the side opposite the angle ADE, is equal to EC opposite EBC; also DE is equal to EB. Therefore,

The two diagonals of any parallelogram, bisect each other. Scholium. The two diagonals of a rhom

bus cut each other at right-angles. For, since in the triangles ABE, BEC, the side AB is equal to BC, AE to CE, and EB is common to both, these two triangles are equal; (Prop. 10;) consequently the angle AEB BEC; hence they are right-angles. (Def. 12.)

D