Cor. If two triangles are mutually equilateral they are equal throughout.

EXERCISES.

1. Prove that if two angles of one triangle are equal to two angles of another, the third angles are equal.

2. Prove that each angle of an equilateral triangle is equal to two-thirds of a right angle.

3. Prove that if a perpendicular be erected on the middle point of a straight line, any point in it will be equally distant from the extremities of that line (Theo. VII).

4. Prove that from a given point without a straight line, only one perpendicular to that line can be drawn. (Cor. 2, Theo. V).

5. If the vertical angle of an isosceles triangle be two-sevenths of a right angle, what will be the value of each of the angles at the base?

SEC. VI.-QUADRILATERALS.

DEFINITIONS.

1. A QUADRILATERAL is a polygon of four sides.

2. A PARALLELOGRAM is a quadrilateral having its opposite sides parallel.

A parallelogram whose angles are all right angles is called a rectangular parallelogram, or simply a RECTANGLE.

A SQUARE is a rectangle whose sides are all equal.

3. A TRAPEZOID is a quadrilateral having only two of its opposite sides parallel.

4. A DIAGONAL of a quadrilateral or other polygon is a straight line joining two angles not adjacent to each other.

5. The AREA of any plane figure is the amount of surface which it contains.

6. Two plane figures are said to be equivalent when their areas are equal.

THEOREM XIII.

The opposite sides and angles of a parallelogram are equal to each other.

A

Let ABCD be a parallelogram. D

It is to be proved that any one of its sides is equal to the side opposite, and any one of its angles to the angle opposite.

B

Draw the diagonal AC. Now, since AC intersects the two parallels AD, BC, the alternate angles DAC, BCA, are equal (Theo. III); also, since it intersects the parallels AB, DC, the ́alternate angles BAC, DCA, are equal. Hence, the two triangles ABC, ADC, have two angles of the one equal to two angles of the other, and the included side AC common; they are,

therefore, equal (Theo. XI); and the side AB is equal to the side DC (Cor., Theo. VII), and the side AD to the side BC, and the angle B to the angle D; also, since the angle DAC is equal to the angle BCA, and the angle BAC to the angle DCA, it follows that the whole angle BAD is equal to the whole angle BCD.

Therefore, the opposite sides and angles, etc.

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

Cor. 2. If ABCD be a rectangular parallelogram, AD and BC will both be perpendicular to each of the parallels AB, DC (Def. 2, Sec. VI). Hence, the perpendicular distance between two parallels is everywhere the same.

Schol. Any side, as AB, of a parallelogram may be taken as its base; and a perpendicular, as CE, let fall A4

from any point in the oppo

site parallel on the base, or the base produced, is called the altitude of the parallelogram.

THEOREM XIV.

If a quadrilateral has two of its sides equal and parallel, it is a parallelogram.

Let the quadrilateral ABCD D have the sides AB, DC, equal and parallel. Then will it be a parallelogram.

A

B

lels AB, DC, the

equal (Theo. III).

we have the side

Join AC; then, since AC intersects the two paralalternate angles BAC, DCA, are Now, in the triangles ABC, ADC, AB equal to the side DC, and the side AC common, and the included angle BAC equal to the included angle DCA; therefore, the two triangles are equal (Theo. VII), and the angle BCA is equal to the angle CAD (Cor., Theo. VII), and these being alternate angles, it follows that AD and BC are parallel (Cor. 1, Theo. III). Hence, ABCD has its opposite sides parallel, and is a parallelogram (Def. 2, Sec. VI).

Therefore, if a quadrilateral, etc.

Schol. If a quadrilateral has its opposite sides equal it is a parallelogram. For the triangles ABC, ADC, will in that case be mutually equilateral, and therefore equal (Cor., Theo. XII); and the parallelism of the opposite sides may be shown by the equality of the alternate angles, as above.

THEOREM XV.

The diagonals of a parallelogram bisect each other. Let ABCD be a parallelogram. It is to be proved that its diagonals, AC, BD, bisect each other in E (Def. 6, Sec. IV).

In the triangles AEB, DEC, since the angle ABD is equal to its alternate angle BDC (Theo. III), and the angle BAC equal to its alternate angle ACD, and the included side AB equal to the included side DC

(Theo. XIII), it follows that the two triangles are equal (Theo. XI); consequently, the side AE is equal to the side EC, and BE to ED (Cor., Theo. VII); that is, the two diagonals bisect each other in E. Therefore, the diagonals, etc.

THEOREM XVI.

The area of a rectangle is equal to the product of its base by its altitude.

Let ABCD be a rectangle.

Ab, bc, etc., taken as the units of length; also, let AD be divided into a certain number of the same units, Af, fg, etc. From b, c, etc., draw straight lines parallel to AD, and from f, etc., draw straight lines parallel to AB. Now, it is evident that the whole rectangle is divided into small squares (Cor. 2, Theo. XIII), each equal to Abef; which may be taken as the unit of area. Of these equal squares there are as many in the tier next to AB as there are units of length in AB; and there are as many equal tiers in the whole figure as there are units of length in AD. Therefore, the whole number of square units in ABCD is equal to the number of linear units in AB multiplied by the number of linear units in AD.

Hence, the area of a rectangle, etc.