double angle VPW must be the central angle of the same arc; i.e. Pis the mid-point between a vertex and orthocentre so, also, are Q and R, similarly. Q. E. D.

A

Z

W

B

FIG. 90.

Def. This remarkable circle is called the 9-point circle, or circle of Feuerbach, of the ▲ ABC.

Corollary. The radius of the 9-point circle is half the radius of the circumcircle.

135. Def. A Polygon all of whose sides touch a circle is said to be circumscribed about it, and the circle is said to be inscribed in the polygon.

Theorem LXXIII. A circle may be inscribed in any Δ.

Proof. Let ABC be any ▲ (see Fig. 59). Draw the inner mid-rays of the angles at A, B, C; they concur in the in-centre I of the A, equidistant from the three sides (why?). About this point as centre with this common distance as radius draw a circle; it will touch the three sides of the A (why and where?). Q. E. D.

N.B. We have seen that the outer mid-rays of the angles concur in pairs with the inner mid-rays of the angles in the three ex-centres E1, E2, E3, also equidistant from the sides

(Fig. 60). The circles about these touch only two sides innerly, but the third side outerly, and hence are called escribed, or ex-circles.

Corollary. Four, and only four, circles touch, each, all the sides of a A.

135 a. Theorem LXXIV.

- In a 4-side circumscribed

about a circle the sums of the two pairs of opposite sides are equal (Fig. 91).

Proof. The sum of the four sides is plainly 2 t + 2 u + 2 v +2w, and the sum of either pair of opposites is t+u+v+w. Q. E. D.

Conversely, If the sums of two pairs of opposite sides of a 4-side be equal, the 4-side is circumscribed about a circle.

Proof. Let two counter sides, AB and DC meet in I, and inscribe a circle K in the triangle ADI. Through B

draw a tangent (Fig. 92) to K at U, and let it cut DI at C'. Then since ABC'D is circumscribed about K, we have

Hence C and C' fall together (why? Art. 56). Q. E. D.

136. Theorem LXXV. The tangent-length from a vertex of a to the in-circle equals half the perimeter of the ▲ less the opposite side (Fig. 93).

Proof.

For the sum of CE + CD + BD + BF is plainly 2a (why?); subtract this from the whole perimeter, a + b + c, and there remains AE+AF = a+b+c − 2 a, or AE = b+c− a = AF. Q.E.D.

2

It is common and convenient to denote the perimeter (Fig. 93) (= measure round = sum of sides) by 2s; then we see that the tangent-lengths from A, B, C, are s— - a, s-b, s-c.

Corollary. The tangent-length from any vertex, A, of a A to the opposite ex-circle and the two adjacent ex-circles are s, s-b, s — c. Hence sa, s, s b, S- - c, are the four tangent-lengths from any vertex, A, of a ▲ to the incircle and the three ex-circles.

These relations are useful and important.

137. Theorem LXXVI. Proof.

For the angle is a continuous magnitude (why?);

hence there are angles of all sizes from zero to a round

I

angle; hence there is an angle, the

n

part of a round angle,

such that, taken n times in addition, the sum will be a round angle. Suppose such an angle drawn, whether or not we can actually draw it, and suppose ʼn such angles placed consecutively around any point O, so as to make a round angle. In other words, suppose ʼn half-rays drawn cutting the round angle about O into n equal angles. Draw a circle about O, with (Fig. 94) any radius, and draw the ʼn chords

subtending the n equal central angles. These chords are all equal (why?), and subtend equal arcs, and they form an n-side. Moreover, the angle between two consecutive sides