A

Cor. II. An angle, ACB, formed by a tangent and chord, has also for its measure one half the arc CB included between its sides. For, drawing the diameter CE, we have ACE a right angle (T. V., C. I.), and B it has for its measure one half of the semicircumference CBE. The angle BCE has

for its measure one half the arc BE; hence, the difference of these angles; that is, the angle ACB has for its measure one half the arc CB.

If we consider the supplementary angle FCB, we have the right angle FCE measured by one half the arc CGE; hence, the sum of the angles FCE and ECB; that is, the angle FCB has for its measure one half the included arc CGEB.

THEOREM XI.

The angle formed by the intersection of two chords is measured by half the sum of the included arcs. And the angle formed by the intersection of two secants is measured by half the difference of the included arcs.

Drawing the chord AF parallel to CD, we have the arc FD= AC (T. VI.), and the angle BAF=BED (B. I., T. XVIII., C.), but the angle BAF is measured, in the case of the chords, by

half the sum of the arcs BD and DF; that is, by half the sum of BD and AC. But in the case of the secants, the angle BAF is measured by half the difference of BD and DF; that is, by half the difference of BD and AC. Consequently, the angle formed by the intersection of two chords is measured by half

the sum of the included arcs; and the angle formed by the intersection of the two secants is measured by half the difference of the included arcs.

Cor. The angle AEC formed by the intersection of two tangents, is measured by half the difference of the concave and convex arcs B AFC and AC, comprehended between the points of contact.

For, drawing AF parallel to CE, we have the arc FC-AC (T. VI.), and the angle BAF=BEC, but the angle BAF is measured

E

A

F

by half the arc AF (T. X., C. II.); that is, by half the difference of AFC and FC, or which is the same, by half the difference of the concave arc AFC and the convex arc AC.

OF INSCRIBED AND CIRCUMSCRIBED POLYGONS.

THEOREM XII.

All triangles are capable of being inscribed in a circle and of circumscribing a circle.

A

E

C

F

G

D

B

First. The three perpendiculars bisecting the three sides of the triangle ABC, meet in the same point G, which is equally distant from A, B, and C (B. I., T. XXXVII.). Hence, if with G as a centre, a circumference be described with a radius equal to the distance from G to either angle of the triangle, it will circumscribe the triangle, and consequently the triangle will be inscribed in the circle. Secondly. The three lines bisecting the three angles of the triangle ABC meet in the same point G, which is equally distant from the three sides of the triangle (B. I., T. XXXVI.). Hence, if with G as a centre, a circumfer

G

A

C

B

ence be described with a radius equal to the distance from G to either side of the triangle, it will be inscribed in the triangle, and consequently the triangle will circumscribe the circle.

Scholium I. Since, in the first case, the perpendiculars can meet in only one point, there can be only one circumference circumscribing a triangle. That is, through three points not in the same straight line, only one circumference can be made to pass. Hence, when two eircumferences have three points common, they must coincide.

If the three points A, B, and C, are all in the same straight line, the perpendiculars drawn bisecting AB, BC and CA will be parallel, and cannot meet, in which case there can be no centre, unless we regard the centre as at an infinite distance.

Scholium II. If, in the second

A

E

F

B

D

case, the sides AB and AC are produced, the point F, where the lines bisecting the exterior angles meet, is also equally distant from the three lines forming the triangle. Therefore, with F as a centre, a new circumference may be described tangent to the three sides. This circle is called the escribed circle. Hence, if lines be drawn bisecting the angles, and the exterior angles of a triangle, they will intersect each other by threes at the centres of the inscribed and escribed circles; thus the three lines which bisect the angles will meet at the same point, giving the centre of the inscribed circle. Any one of the lines bisecting an angle of the triangle will intersect, at the same point, two of the lines which bisect the exterior angles, giving the centre of an escribed circle.

Scholium III. This Theorem, taken in connection with Theorem XXXV., Book First, shows that, when the triangle is right-angled at C, the centre of the circumscribed circle will be at D, the middle point

of the hypotenuse, for in this case the three

lines DA, DB, and DC are all equal.

A

C

D

B

If the triangle is isosceles, the centre of the circumscribing circle as well as that of the inscribed circle will be in the line bisecting the angle formed by the equal sides.

If the triangle is equilateral, the centres of the two circles will coincide, and the circumscribed and inscribed circles will then be concentric.

THEOREM XIII.

In any inscribed quadrilateral, the sum of the opposite angles is equal to two right angles.

The angle at B is measured by half the arc CDA, and the opposite angle at D is measured by half the arc ABC; hence the sum of the angles at B and D is measured by half the entire circumference, which is the measure of two right angles.

First

A

B

A

C

B

C

D

D

D

E

Cor. Conversely, a quadrilateral may be inscribed when the sum of its opposite angles is equal to two right angles. If we describe a circumference through the three points A, B, and C, which can always be done (T. XII., S. I.), and it does not pass through D, then the point D must be either within or without this circumference. suppose it within. Produce AD until it meet the circumference at E, then; by the Theorem itself, we shall have the sum of ABC and CEA =2 right angles: but by hypothesis ABC+CDA=2 right angles; consequently CEA = CDA, which is absurd (B. I., T. VI., C.). Hence, the point D cannot be within the circumference. By a similar process, we can show that it cannot be without the circumference. It must therefore be in the circumference.

Scholium. The rectangle, which includes the square, is the only parallelogram which can be inscribed in a circle. The diagonals of the rectangle are diameters of the circumscribing circle.

EGB, will be equal, since they have the hypotenuse and a side of the one respectively equal to the hypotenuse and a side of the other (B. I., T. XXVI.), and BF BG. In a similar manner

DH=DK. Conse

we may show that AF AK; CH=CG; DH = DK.
AF-AK;
quently by adding, we have AB+CD = BC+AD.

Cor. When the sums of the opposite sides of a quadrilateral are equal, it is capable of circumscribing a circle.

For, if we describe a circum

ference tangent to AD, AB, and BC, which can always be done (T. XII., S. I. or II.), it will also be tangent to DC. For, if the side DC is not tangent to the circumference, it must be either a secant or lie wholly without the circumference. We will first suppose it to be a secant. If DE

is drawn tangent to this circumference, we shall have AD+BE = AB+DE, but by hypothesis we have AD+BC=AB+ DC; hence, by subtraction, we obtain EC=DE - DC; that is, one side of a triangle is equal to the difference of the other two sides, which is impossible (B. I., T. VII.). Hence the side DC cannot be secant. In a similar way we can show that it cannot be wholly without the circumference. It must therefore be tangent. Scholium. The rhombus and the square are the only quadrilaterals capable of circumscribing a circle.

THEOREM XV.

All regular polygons are capable of being inscribed in a circle, and of circumscribing a circle.

Let ABCDEF be any regular polygon;

B

and through three consecutive angles A, B, and C, describe the circumference of a circle (T. XII.). This circumference will also pass through all the other angles of the polygon. For, drawing from the centre G, the lines GA, GB, GC, and GD, we shall form three triangles, GAB, GBC,