Of Polygons.

Let A, B, C, D, and E, be the outward angles of a polygon formed by producing

all the sides.

Then will

A+B+C+D+E=four right angles.

E

a

B

For, each interior angle, plus its exterior angle, as A+a, is equal to two right angles (Th. ii). But there are as many exterior as interior angles, and as many of each as there are sides of the polygon: hence, the sum of all the interior and exterior angles will be equal to twice as many right angles as the polygon has sides.

But the sum of all the interior angles together with four right angles, is equal to twice as many right angles as the polygon has sides (Th. xxi): that is, equal to the sum of all the inward and outward angles taken together.

From each of these equal sums take away the inward angles, and there will remain, the outward angles equal to four right angles (Ax. 3).

THEOREM XXIII

The opposite sides and angles of every parallelogram are equal, each to each: and a diagonal divides the parallelogram into two equal triangles.

Let ABCD be any parallelogram, and DB a diagonal: then will the opposite sides and angles be equal to each other, each to each, and the diagonal DB will divide the parallelogram into two equal triangles.

B

For, since the figure is a parallelogram, the sides AB, DC are parallel, as also the sides AD, BC. Now, since the

Of Parallelograms.

parallels are cut by the diagonal DB, the alternate angles will be equal (Th. xii): that is the angle

ADB DBC

and

BDC=ABD.

Hence the two triangles ADB BDC, having two angles in the one equal to two angles in the other, will have their third angles cqual (Th. xvii. Cor. 1), viz. the angle A equal to the angle C, and these are two of the opposite angles of the parallelogram.

Also, if to the equal angles ADB, DBC, we add the equals BDC, ABD, the sums will be equal (Ax. 2): viz. the whole angle ADC to the whole angle ABC, and these are the other two opposite angles of the parallelogram.

Again, since the two triangles ADB, DBC, have the side DB common, and the two adjacent angles in the one equal to the two adjacent angles in the other, each to each, the two triangles will be equal (Th. v): hence, the diagonal divides the parallelogram into two equal triangles.

Cor. 1. If one angle of a parallelogram be a right angle, each of the angles will also be a right angle, and the parallelogram will be a rectangle.

Cor. 2. Hence, also, the sum of either two adjacent angles of a parallelogram, will be equal to two right angles.

THEOREM XXIV.

If the opposite sides of a quadrilateral, are equal, each to each, the equal sides will be parallel, and the figure will be a pa rallelogram.

Of Parallelegrams.

Let ABCD be a quadrilateral, having its opposite sides respectively equal, viz.

then will these sides be parallel, and the figure will be a parallelogram.

For, draw the diagonal BD. Then, the two triangles ABD BDC, have all the sides of the one equal to all the sides of the other, each to each: therefore, the two triangles are equal (Th. viii); hence, the angle ADB, opposite the side AB, is equal to the angle DBC opposite the side DC; therefore, the sides AD, BC, are parallel (Th. xiii). For a like reason DC is parallel to AB, and the figure ABCD is a parallelogram.

THEOREM XXV.

If two opposite sides of a quadrilateral are equal and parallel, 'he remaining sides will also be equal and parallel, and the figure mill be a parallelogram.

Let ABCD be a quadrilateral, having D the sides AB, CD, equal and parallel: then will the figure be a parallelogram.

For, draw the diagonal DB, dividing

the quadrilateral into two triangles. Then,

since AB is parallel to DC, the alternate angles, ABD and BDC are equal (Th. xii): moreover, the side BD is common: hence the two triangles have two sides and the included ang. of the one, equal to two sides and the included angle of the other: the triangles are therefore equal, and consequently AD is equal to BC, and the angle ADB to the angle DBC and consequently, AD is also parallel to BC (Th xiii Therefore, the figure ABCD is a parallelogram.

Of Parallelograms.

THEOREM XXVI.

The two diagonals of a parallelogram divide each other into equa parts, or mutually bisect each other.

Le ABCD be a parallelogram, and AC, BD its two diagonals intersecting at

E. Then will

AE-EC

and BE-ED.

Comparing the two triangles AED and

BEC, we find the side_AD=BC (Th. xxiii), the angle ADE EBC and EAD=ECB: hence, the two triangles are equal (Th. v): therefore, AE, the side opposite ADE, is equal to EC, the side opposite EBC; and ED is equal to EB

D

E

Sch. In the case of a rhombus (Def. 48), the sides AB, BC being equal, the triangles AEB and BEC have all the sides of the one equal to the corresponding sides of the other, and are therefore equal.A Whence it follows that the angles AEB and BEC are equal. Therefore, the diagonals of a rhombu bisect each other at right angles.

1. THE circumference of a circle is a curve line, all the points of which are equally distant from a certain point within called the centre.

2. The circle is the space bounded by this curve line.

3. Every straight line, CA, CD, CE, drawn from the centre to the circumference, is called a radius or semidiameter. Every line which, like AB, passes through the centre and terminates in the circumference, is called a diameter.

4. Any portion of the circumference, as EFG, is called an arc.

5. A straight line, as EG, joining the E extremities of an arc, is called a chord.

6 A segment is the surface or portion of a circle included between an arc and its chord. Thus EFG is a segment.

E

B