VI. Problems.

PROBLEM III.

412. To bisect a given arc.

GIVEN, the arc BD.

REQUIRED, to bisect it.

CONSTRUCTION. Join BD, and bisect the sect BD in F; at F erect a perpendicular cutting the arc in C.

C is the mid point of the arc.

PROOF. Join BC,CD.

▲ BCF≈ ▲ DFC,

(124. Triangles having two sides and the included angle in each equal are con

(374. In the same circle, equal chords subtend equal minor arcs.)

413. A polygon is said to be circumscribed about a circle when all its sides are tangents to the circle.

The circle is then said to be inscribed in the polygon. 414. A circle which touches one side of a triangle and the

other two sides produced is called an Escribed Circle.

PROBLEM IV.

415. To describe a circle touching three given lines which are not all parallel, and do not all pass through the same point.

GIVEN, three lines intersecting in A, B, and C.
REQUIRED, to describe a circle touching them.

CONSTRUCTION. Draw the bisectors of the angles at A and C.
These four bisectors will intersect in four points, 0, 01, 02, 03.

A circle described with any one of these points as center, and its perpendicular on any one of the three given lines as radius, will touch all three.

PROOF. By 186, every point in the bisector of an angle is equally distant from its arms;

Therefore, since O is on the bisector of A, the perpendicular from O on AB equals the perpendicular from O on AC, which also equals the perpendicular from O on CB, since O is also on the bisector of C.

416. The four tangents common to two circles occur in two pairs intersecting on the common axis of symmetry.

PROBLEM V.

417. In a given circle to inscribe a triangle equiangular to a given triangle.

REQUIRED, to describe in the © a ▲ equiangular to ▲ ABC. CONSTRUCTION. Draw a tangent GH touching the circle at the

point D.

Make

HDK = C, and GDL = X A.

K and L being on the circumference, join KL.

DKL is the required triangle.

PROOF. KDL = B.

(174. The three angles of a triangle are equal to a straight angle.)

(375. An inscribed angle is half the angle at the center on the intercepted arc.) (397. An angle formed by a tangent and a chord is half the angle at the center

on the intercepted arc.)

BOOK IV.

REGULAR POLYGONS.

I. Partition of a Perigon.

PROBLEM I.

418. To bisect a perigon.

SOLUTION. To bisect the perigon at the point O, draw any line through O.

This divides the perigon into two straight angles, and all straight angles are equal.

419. COROLLARY. By drawing a second line through at right angles to the first, we cut the perigon into four equal parts; and as we can bisect any angle, so we can cut the perigon into 8, 16, 32, 64, etc., equal parts.

PROBLEM II.

420. To trisect a perigon.

SOLUTION. To trisect the perigon at the point O, to O draw any line BO; on BO produced take a sect OC; on OC construct, by 132, an equilateral triangle CDO.

DOB is one-third of a perigon.

For DOC is one-third of a st. 4,

(174. The three angles of a triangle are equal to a straight angle.)

421. COROLLARY. Since we can bisect any angle, so we may cut the perigon into 6, 12, 24, 48, etc., equal parts.

blem

422. REMARK. To trisect any given angle is a pro beyond the power of strict Elementary Geometry, which allows the use of only the compasses and an unmarked ruler. There is an easy solution of it, which oversteps these limits only by using two marks on the straight-edge. The trisection of the angle, the duplication of the cube, and the quadrature of the circle, are the three famous problems of antiquity.