OCB = OCD, and the sides which contain these equal angles are equal to one another, each to each, therefore OD= OB, and ODC= OBC. But OBC is half of ABC, that is of CDE, since the polygon is equiangular. Therefore ODC is half of CDE, and therefore OD bisects the angle CDE.

In the same manner it may be shewn that OE = OC, and bisects the angle at E. Hence O is the centre of the circle circumscribed about the polygon, of which OB is the radius.

And because AB, BC, CD ... are equal chords in this circle, of which O is the centre, therefore the perpendiculars OP, OQ, OR on those chords are all equal; and a circle described with centre O and radius OP will pass through Q, R... and touch AB, BC, CD ... in the points P, Q, R... That is O is the centre of the inscribed circle of which OP is the radius.

PROBLEM 8.

To construct a regular polygon of four, eight, sixteen ... sides, and inscribe them in, or describe them about a given circle.

F

Take the centre of the given circle, and draw through it two diameters at right angles to one another, meeting the circumference in A, B, C, D; then by Theorem 12, if AB, BC... be joined, we get a regular inscribed polygon of four sides; and if tangents be drawn at A, B, C and D, we get a regular circumscribed polygon of four sides, that is a square.

D

E

To describe an octagon, bisect (by Part 1. Prob. 1) the angles AOB, BOC, COD, DOA by lines meeting the circle in E, F, G, H; then A, E, B, F... are the angular points of a regular inscribed octagon, or the points of contact of the sides of a regular circumscribed octagon.

Similarly, by again bisecting the angles at O, a regular sixteen-sided polygon may be constructed; and hence a thirty-two-sided figure, and so generally a polygon of 2 sides, may be constructed, when n is any integer greater than I.

PROBLEM 9.

To construct regular polygons of three, six, twelve... sides, and inscribe them in, or describe them about a given circle. Let Ở be the centre of the circle. On OA, one of the

B

4

radii, make an equilateral triangle AOP, and take AOB, double of the angle AOP.

Then since each angle of an equilateral triangle is onethird of two right angles; therefore its double is one-third of four right angles; and therefore two other equal angles BOC, COA will fill up the space round O.

Hence the angles at O being equal, A, B, C are the angular points of a regular inscribed polygon of three sides.

As before, by bisecting the angles at O we obtain the angular points of the regular hexagon; and by bisecting these angles we obtain the dodecagon; and so generally a regular polygon of 3 x 2" sides may be constructed, where n may have any integral value, including zero.

The Theorems at present proved do not enable the student to construct any regular polygons except those included in the foregoing problems.

Hereafter we shall shew that a regular pentagon can be described, and by means of it the decagon, quindecagon, &c.

THEOREM.

The area of a circle is equal to half the rectangle contained by the radius and a straight line equal to the circumference.

Let C be the centre of a circle DEF. And let AB be the side of a polygon described about the circle, CD a radius drawn to the point of contact.

E

B

Then the triangle ABC= the rectangle contained by CD and AB.

And since the whole polygon can be divided into triangles by lines drawn from the angular points to the centre,

=

The area of the polygon the rectangle contained by the radius of the circle and perimeter of the polygon.

Now this is true whatever may be the number of sides of the polygon.

But by perpetually increasing the number of sides of the polygon, that polygon approaches nearer and nearer to the circle.

So finally the area of the circle = rectangle contained by radius and circumference.

Remark. This will give the area of the circle when we know the lengths of the radius and the circumference. The length of the latter cannot be found by the use of the rule and compasses. But if we added to our mathematical instruments a cylinder that might be rolled on the paper, or a tape that might be unrolled from a cylinder, we should have that length at once, and consequently the area.

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MISCELLANEOUS THEOREMS AND PROBLEMS.

I.

Prove that the two tangents drawn to a circle from any external point are equal.

2. If from a point without a circle two tangents AB, AC are drawn, the chord of contact BC will be bisected at right angles by the line from A to the centre.

3. If a circle is inscribed in a right-angled triangle, the excess of the two sides over the hypothenuse is equal to the diameter of the circle.

4. If a quadrilateral figure be described about a circle, the sums of the opposite sides will be equal to one another.

5. If a six-sided figure be circumscribed about a circle, the sums of the alternate sides will be equal.

6. If a quadrilateral figure be described about a circle, the angles subtended at the centre by any two opposite sides are together equal to two right angles.

7. AOB, COD are two chords of a circle at right angles to one another; prove that the squares of OA, OB, OC, and OD, are together equal to the square of the diameter.