PROPOSITION XV. THEOREM.

The angle formed by a tangent, and a chord drawn from the point of contact, is equal to an angle in the alternate segment of the circle; that is, to an inscribed angle subtended by the arc intercepted by the sides of the former.

Let ABC touch the circle BDEF in B, making with the chord BD the angles ABD, CBD; the former will be equal to an angle in the segment BGD, and the latter to an angle in the segment BFEĎ.

A

For, draw the diameter BE and the chord ED; draw also EG, BG, and DG to any point G in the arc BGD; then, ABE being a right angle (Prop. IX.), it is equal to the angle BGE in F the semicircle (Prop. XIV. Cor. 3.), and the angle EBD is equal to the angle EGD, for it is subtended by the same arc ED (Prop. XIV. Cor. 2.); therefore the whole angle ABD is equal to

the whole angle BGD in the alternate segment.

B

G

E

Again, the angle CBD, together with DBE, make a right angle, also the angle DEB, together with DBE, make a right angle; hence the angle CBD is equal to the angle DEB in the alternate segment

Cor. 1. An angle in a segment greater than a semicircle is acute, and an angle in a segment less than a semicircle is obtuse.

Cor. 2. Therefore the segment which contains a right angle must be a semicircle, the segment which contains an obtuse angle must be less, and that which contains an acute angle must be greater than a semicircle.

Scholium.

This proposition and the preceding may both evidently be combined in the following general enunciation:

An inscribed angle, and an angle formed by a tangent and a chord, are each equal to an angle at the centre, subtended by half the arc included by their sides. The following proposition is the converse of this.

PROPOSITION XVI. THEOREM. (Converse of Props. XIV. and XV.)

If an angle, whose sides include an arc of a circle, be equal to an angle at the centre whose sides include half that arc, the vertex of the former shall be in the circumference; that is, it shall be either an inscribed angle, or contained by a tangent and a chord.

Let the sides of the angle ADB include the arc AB, and let it be equal to an angle at the centre of the circle, whose sides include half that arc; the point D shall be in the circumference.

For, let D be supposed to lie either within, or without, the circumference, and in the former case let C be the point where the production of one of the sides cuts the circumference, and in the latter case let C be a point where one of the sides meets the circumference, and let CE be drawn parallel to the other side DB.

Then, however the sides of the angle ACE be situated with regard to the circumference; that is to say, whether both are chords, or one a chord and the other a tangent, this angle will in either case be equal to an angle at the centre, subtended by half the arc included by its sides (Prop. XV. Scholium); but, by hypothesis, the angle ADB, which is, by construction, equal to ACE (Prop. XIV. Cor. 1. B. I.), is equal to an angle at the centre, subtended by half the arc included by its sides, but these included arcs are unequal; their halves are, therefore, unequal, and yet they subtend equal angles at the centre, which is impossible (Prop. IV.). The point D, therefore, can lie neither within, nor without, the circumference; it is, therefore, in it.

A

D

ED

B

F

B

B

E

B

Cor. 1. In the case where D is within the circle, the angle ADB is equal to an inscribed angle C, subtended by an arc, equal to the sum of the arcs AB, CF (Prop. X.), that is, an angle formed by the intersection of two chords, is equal to an inscribed angle subtended by the sum of the opposite intercepted

arcs.

Cor: 2. In the case where D is without the circle, the angle ADB is equal to an inscribed angle, subtended by an arc equal to the difference of the intercepted arcs, that is, an angle whose vertex is without the circle, and whose sides meet the circumference, is equal to an inscribed ungle subtended by the difference of the opposite intercepted arcs.

The opposite angles of a quadrilateral inscribed in a circle, are together equal to two right angles. The opposite angles A, C, or ABC, ADC, of the inscribed quadrilateral ABCD, are together equal to two right angles. For let EBF be a tangent to the circle

at B, and join BD, then the angle DBF is equal to the angle A, and DBE to the angle C (Prop. XV.); but the angles DBF, DBE are together equal to two right angles: therefore the angles A, C are together equal to two right angles; and since the four angles of the quadrilateral are together equal to four right angles (Prop. XVII. Cor. 1. B. I.), the remaining two must be equal to two right angles.

E

F

Cor. 1. If one side of an inscribed quadrilateral be produced, the exterior angle will be equal to the interior opposite

one.

Cor. 2. A quadrilateral, of which the opposite angles are not equal to two right, cannot be inscribed in a circle.

PROPOSITION XVIII. THEOREM. (Converse of Prop. XVII.) If the opposite angles of a quadrilateral be together equal to two right angles, a circle may be circumscribed about it.

Let ABCD be a quadrilateral, the opposite angles B, D

of which are together equal to two right angles, a circle may be circumscribed about it, that is, the circumference which passes through the three points A, B, C, shall also pass through D.

D

For if D were to lie within the circle, the angle D would be greater than if it were in the circumference (Prop. XVI. Cor. 1.), and consequently the angles B, D would be together greater than two right angles (Prop. XVII.); and if D were to lie without the circle the angle D would be less than if it were in the circumference, and therefore the angles B, D would, in this case, be less than two right angles; D therefore can lie neither within nor without the circle, that is, it is in the circumference.

B

Cor. 1. If two opposite angles of a quadrilateral be together equal to the other two opposite angles, a circle may be described about the quadrilateral (Prop. XVII. Cor. 3. B. I.).

Cor. 2. A trapezium may be inscribed in a circle, provided the non-parallel sides are equal.

For if the non-parallel sides AD, BC, of the trapezium ABCD, be equal, and perpendiculars DE, CF, be drawn to AB, the triangles ADE,

BCF will be equal, the angle A A E equal to the angle B, and the angle

F

B

ADE to the angle BCF, and consequently the angle ADC must be equal to the angle BCD, so that two opposite angles of the trapezium are equal to the other two, therefore a circle may be described about it (Cor. 1.).

BOOK IV.

POSTULATE.

From any point as a centre with any radius, a circumference may be described.

PROPOSITION I. PROBLEM.

To divide a given straight line AB, into two equal parts.

From the points A and B as centres, with any radius greater than half AB, describe two arcs, which must necessarily cut each other (Prop. XIII.. Schol. 4. B. III.); draw the straight Aline CD through the points of intersection, and it will pass through M, the middle of AB; for CD is perpendicular to AB (Prop. XIII. Cor. 3. B. III.),

M

-B

and the points A, B, are equally distant from C, therefore they are equally distant from M (Prop. XXII. Schol. B. I.).

Cor. CD not only divides AB into two equal parts, but it is at the same time perpendicular to AB.

PROPOSITION II. PROBLEM.

From a given point P, in a straight line AB, to draw a perpendicular to that line.

In the given straight line, or in its prolongation, take two points C D, equally distant from P; and from these points as centres, with a radius longer than CP, describe arcs which will intersect in E; draw PE and it will be the perpendicular required; for it is drawn

A

P

D