DEF. 17. If all the sides of a rectilineal figure touch a circle lying within the figure, the circle is said to be inscribed in the figure, and the figure to be circumscribed about the circle.

PROB. 9. About a given circle to circumscribe a triangle equiangular to a given triangle.

Let ABC be the given circle, DEF the given triangle:

it is required to circumscribe about ABC a triangle equiangular to DEF.

Find the centre O, and draw any diameter AG; make the angle GOC equal to the angle DEF, and on the other side of AG make the angle GOB equal to the angle DFE;

III. Prob. 2, and I. Prob. 5.

draw tangents to the circle at the points A, B, C, III. 20, Cor. 5.

forming the triangle HKL:

HKL shall be the triangle required.

Because the angles at A and C are right angles,

III. 20, Cor. 2.

therefore a circle may be described about the quadrilateral

CKAO;
therefore the angle CKA is equal to the exterior angle COG,

and therefore to the angle DEF.

III. 19, Cor. 2.

III. 19, Cor. 1.

Constr.

In like manner the angle ALB is equal to the angle DFE; hence the remaining angle KHL is equal to the remaining angle EDF,

and the triangle HKL is equiangular to the triangle DEF, and it is circumscribed about the circle ABC.

I. 25.

Q.E.F.

Ex. 115. About a given circle to circumscribe a quadrilateral equiangular to a given quadrilateral,

THEOR. 26. If the whole circumference of a circle is divided into any number of equal arcs, the inscribed polygon formed by the chords of these arcs is regular; and the circumscribed polygon formed by tangents drawn at all the points of division is also regular.

Let the whole circumference of the circle ABC be divided into any number (say five as in the figure) of equal arcs at the points A, B, C. . ., and let the chords AB, BC, CD, . . . be drawn:

Again, each angle of the polygon ABCD ... stands upon an arc which is less than the whole circumference by two of the arcs AB, BC, CD . . ., but the arcs AB, BC, CD,.

equal;

are all Hyp.

therefore the remaining arcs on which the angles of the polygon

Next, let tangents be drawn at the points A, B, C, D, SO as to form a polygon PQRS III. 20, Cor. 5. then shall the circumscribed polygon PQRS . . . be regular.

:

III. Prob. 2.

Hyp.

Find the centre O, and join OA, OB, OC. Rotate the figure OAPB round O until A coincides with B : then because the arc AB is equal to the arc BC,

therefore B will fall on C,

and the tangents AP and BP will coincide with the tangents BQ and CQ, since each tangent is at right angles to the radius drawn

to its point of contact ;

III. 20, Cor. 2.

therefore the figure OAPB coincides with the figure OBQC;

and therefore the polygon PQRS . . . is equilateral, since each of its sides is made up of two of these tangents.

Again, the angle APB coincides with the angle BQC, and is therefore equal to it;

that is, any angle of the polygon is equal to the next succeeding angle;

therefore the polygon is equiangular.

Hence the polygon is regular.

I. Def. 23.

Q.E.D.

Ex. 116. Any equilateral figure inscribed in a circle is also equiangular.

Ex. 117. In any equilateral figure inscribed in a circle each side is equal to the next side but one; and hence, when the number of sides is odd, the figure is equilateral.

THEOR. 27. The bisectors of the angles of a regular polygon meet in a point which is equidistant from all the vertices of the polygon and from all the sides.

Let ABCD . . . be a regular polygon, and let the angles of the polygon at A and B be bisected by straight lines meeting at Q; join OC, OD

.

then shall OC, OD... bisect the angles of the polygon ai

C, D . . .

In the triangles OBA, OBC,

the side AB is equal to the side BC,

the side OB is common to both,

and the angle OBA is equal to the angle OBC;

therefore the angle OAB is equal to the angle OCB.

But the angle OAB is half the angle of the polygon at A,

and the angle at A is equal to the angle at C;

Hyp.

Нур.

I. 5.

Hyp.

Hyp.

therefore the angle OCB is half the angle at C,

that is, OC bisects the angle at C.

In like manner OD . . . bisects the other angles of the polygon.

Again, because the angles of the polygon at A and B are

equal,

Hyp.

therefore the angles OAB, OBA, the halves of these, are also equal;

therefore OA is equal to OB.

Ax. h.

I. 7.

In like manner OB is equal to OC, and so on.

Again, draw OK, OL, OM... perpendicular to the sides