COR. 2. Every angle, BAD, inscribed in a semi-circle, is a right angle; for it is measured by the half of the semi-circumference BOD, or the fourth part of the circumference. The same thing may be shown in another way by a reference to the Theorem 33, Exercises on Book I.

B

COR. 3. Every angle, BAC (See Fig., Cor. 1), inscribed in a segment greater than a semi-circle is an acute angle, for it is measured by the half of the arc BOC, less than a semi-circumference. And every angle, BOC, inscribed in a segment less than a semi-circle is an obtuse angle; for it is measured by the half of the arc BAC, greater than a semi-circumference.

B

COR. 4. The opposite angles, A and C, of an inscribed quadrilateral, ABCD, are together equal to two right angles; for the angle BAD is measured by half the arc BCD; and the angle BCD is measured by half the arc BAD; hence, the two angles BAD, BCD, taken together, are measured by half the circumference; hence, their sum is equal to two right angles.

This corollary may also be conveniently expressed as follows: Every angle, BAC (See Fig.,

D

Cor. 1), inscribed in one of the segments determined by the chord BC, is the supplement of any angle, BOC, inscribed in the other segment determined by the same chord, BC.

PROPOSITION XXI.

THEOREM.

The angle, BAC, formed by a tangent and a chord, is measured by half the arc, AMDC, included between its sides.

M

D

C

From A, the point of contact, draw the diameter AD. The angle BAD is right (Prop. IX.), and is measured by half the semi-circumference, AMD; the angle DAC is measured by the half of DC; hence, BAD + DAC, or BAC, is measured by the half of AMD plus the half of DC, or by half the whole arc, AMDC. In like manner it may be shown that the angle CAE is measured by half the arc, AC, included between its sides.

B

A

E

PROPOSITION XXII.

THEOREM.

The angle, AOB, whose vertex, O, is within the circumference, is measured by half the arc, AB, included between its sides, plus half the arc, DC, included between their prolongations.

For, join DB. The angle, AOB, exterior

to the triangle, ODB, is equal to the sum of D

The angle, AOB, whose vertex is without the circumference, and whose sides are secants, is measured by half the greater arc, AB, intercepted between its sides, minus half the less arc, CD.

Join CB. The angle ACB, exterior to the triangle OCB, is equal to the sum of the interior angles, O and B. Then the angle O is equal to the difference of the angles ACB and B. But the angle ACB is measured by half of AB, and the angle B by half of CD. Hence, the angle O is measured by the half of AB minus the half of CD.

D

B

A

SCHOLIUM. This proposition is still true when one of the sides of the angle is tangent to the circumference, also when both of them are tangents; and the demonstration is the same.

COR. The arc, BAC, of the circumference, in which the angle CDB is inscribed, is the locus of the vertices of the angles equal to CDB, whose sides pass through C and B, and which lie on the same side of CB with D.

D

A

G

M

For, first, every angle inscribed in BAC has the same measure as CDB, that is, half the arc CB; second, every angle, CMB, whose sides pass through C and B, and whose vertex, M, lies within the segment, has a greater measure than half the arc CB (Prop. XXII.); and, third, every angle CGB, whose sides pass through C and B, and whose vertex lies without the arc BAC, has a less measure than half the arc CB (Prop. XXIII.).

B

PROPOSITION XXIV.

THEOREM.

If two opposite angles, ADC, ABC, of a quadrilateral, are together equal to two right angles, this quadrilateral may be inscribed in a circle.

B

M

C

Describe a circumference through the three points A, D, C, and join AC. The angle ADC will be measured by half the arc AMC; hence, the angle ABC, being the supplement of ADC, is equal to any one of the angles inscribed in the segment AMC. Hence, its vertex, B, must be on the arc AMC (Prop. XXIII., Cor.). Therefore, the circumference passing through A, D, and C, passes also through B.

COR. If the sum of two opposite angles of a quadrilateral is greater or less than two right angles, the quadrilateral cannot be inscribed in a circle.

PROBLEMS RELATING TO THE FIRST TWO BOOKS.

PROBLEM I.

To divide the given straight line AB into two equal parts.

From the points A and B, as centres, with a radius greater than the half of AB, describe two arcs cutting each other in D. will be equally distant from the points A and B.

The point D Find, in like

manner, above or beneath the line AB, a second

point, E, equally distant from the points A and B ; through the two points D, E, draw the line DE: it will bisect the line AB at the point C.

AX

For the two points D and E, being each equally distant from the extremities A and B, must both be A in the perpendicular raised from the middle of AB (Book I., Prop. XVIII., Cor. 2). But through two given points only one straight line can pass; hence, the line DE will itself be that perpendicular which divides the line AB into two equal parts at the.point C.

ЖЕ

SCHOLIUM. The problem, On a given line, AB, as diameter, describe a circle, depends immediately on the above. For the first step is to find the middle point of the line AB, which is the centre of the circle.

PROBLEM II.

At a given point, A, in the line BC, to erect a perpendicular to this line. Take two points, B and C, at equal distances from A; then, from the points B and C, as centres, with a radius greater than BA, describe two arcs cutting each other in D; draw AD, and it will be the perpendicular required.

B

A

For the point D, being equally distant from B and from C, belongs to the perpendicular erected at the middle of BC; therefore, AD is that perpendicular.

SCHOLIUM. If the point, P, were the extremity of the line, and if the line could not be produced beyond it, then

a different construction must be employed.

Thus, from any point, C, taken without the line, with a radius equal to the distance CP, describe a circumference, and from D, where it cuts AP, draw the diameter DE, and join EP; it will be the perpendicular required (Prop. XX., Cor. 2).

E

P

PROBLEM III.

From a given point, A, without the straight line BD, to let fall a perpendicular to this line.

From the point A, as a centre, and with a radius sufficiently great, describe an arc, cutting the line BD at the two points, B and D; then mark a point, E, equally distant from the points B and D, and draw AE; it will be the perpendicular required.

For the two points A and E are each equally distant from the points B and D ; therefore, the line AE is perpendicular to BD, at its middle point.

B

SCHOLIUM. If the point P were opposite the extremity of the line AB, or nearly so, and if AB could not be produced beyond this extremity, the following construction may be employed.

From any point, A, of AB, with a radius equal to its distance from P, describe an arc; then, from a second point in AB, with its distance from P as a radius, describe another arc. Join P with the other point of intersection of these arcs, and this line will be the perpendicular required (Prop. XI., Cor. 1).

A.

B

PROBLEM IV.

At the point A in the line AB, to make an angle equal to the given angle, K.

From the vertex, K, as a centre, and with any radius, describe the arc IL, terminating in the two sides of the angle; from the point A, as a centre, and with a radius, AB, equal to KI, describe the indefinite arc BO; then take a radius equal to the chord LI; with which, from the point B, as a centre, describe an arc cutting the indefinite arc

K

BO in D; draw AD, and the angle DAB will be equal to the given angle K.

For the two arcs, BD, LI, have equal radii and equal chords; therefore, they are equal (Prop. IV.); therefore, the angles BAD, IKL, measured by them, are equal.