them by arcs of a circle, on account of the facility with which arcs can be made equal to given arcs, and for various other reasons. At all events, if the measurement of angles by arcs of a circle is in any degree indirect, it is still equally easy to obtain the direct and absolute measure by this method; since, on comparing the arc which serves as a measure to any angle, with the fourth part of the circumference, we find the ratio of the given angle to a right angle, which is the absolute measure.

125. Scholium 2. All that has been demonstrated in the last three propositions, concerning the comparison of angles with arcs, holds true equally, if applied to the comparison of sectors with arcs; for sectors are not only equal when their angles are so, but in all respects proportional to their angles; hence two sectors ACB, ACD, taken in the same circle, or in equal circles, are to each other as the arcs AB, AD, the bases of those sectors. It is hence evident that the arcs of the circle, which serve as a measure of the different angles, may also serve as a measure of the different sectors, in the same circle, or equal circles.

126. An inscribed angle is measured by half the arc, included between its sides.

Let BAD be an inscribed angle, and let us first suppose that the centre of the circle lies within the angle BAD. Draw the diameter AE, and the radii CB, CD.

E

The angle BCE, being exterior to the triangle ABC, is equal to the sum of the two interior angles CAB, ABC (78.): but the triangle BAC being isosceles, the angle CAB is equal to ABC; hence the angle BCE is double of BAC. Since BCE lies at the centre, it is measured by the arc BE; hence BAC will be measured by the half of BE. For a like reason, the angle CAD will be measured by the half of ED; hence, BAC+CAD, or BAD will be measured by the half of BE+ED, or of BD.

Suppose, in the second place, that C the centre lies without the angle BAD. Then, drawing the diameter AE, the angle BAE will be measured by the half of BE; the angle DAE by the half of DE: hence their difference BAD will be measured by the half of BE minus the half of ED, by the half of BD.

or B

Hence every inscribed angle is measured by the half of the arc included between its sides.

127. Cor. 1. All the angles BAC, BDC, inscribed in the same segment are equal; because they are all measured by the half of the same arc BOC.

B

E

128.

Cor. 2. Every angle BAD, inscribed in a semicircle, is a right angle; because it is measured by half the semicircumference BOD, that is, by the fourth part of the whole circumference.

To demonstrate the same property another

way, draw the radius AC: the triangle BAC is isosceles, hence the angle BAC=ABC; the triangle CAD is also isosceles, hence the angle CAD=ADC; hence BAC+CAD, or BAD=ABD+ADB. But if the two angles B and D of the triangle ABD are together equal to the third BAD, all the three angles will be together equal to twice BAD; we already know that they are equal to two right angles; therefore, BAD is equal to one.

129. Cor. 3. Every angle BAC (see the diagram of 127.) inscribed in a segment greater than a semicircle, is an acute angle; for it is measured by the half of the arc BOC, less than a semicircumference.

And every angle BOC, inscribed in a segment less than a semicircle, is an obtuse angle; for it is measured by the half of the arc BAC, greater than a semicircumference.

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

131.

D

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

Let BE be the tangent and AC the chord. From A, the point of contact, draw the diameter AD. The angle BAD is right (110.) and is measured by half the semicircumference 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.

M

B

D

A

E

It might be shewn, in the same manner, that the angle CAE is measured by half the arc AC included between its sides.

PROBLEMS RELATING TO THE FIRST TWO BOOKS.

PROBLEM.

132. To divide a given straight line into two equal parts.

Let AB be the given straight line.

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; the point D will be equally distant from A and B. 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 and E, draw the line DE: it will bisect the line AB in C.

D

C 함

же

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

PROBLEM.

133. At a given point, in a given straight line, to erect a perpendicular to this line.

Let A be the given point, and BC the given line.

A

Take the 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 intersecting each other in D; draw AD: it will be the perpendicular required.

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

134. Scholium. The same construction serves for making a right angle BAD, at a given point, A, on a given straight line BC.

PROBLEM.

135. From a given point, without a straight line, to let fali a perpendicular on this line.

Let A be the point, and BD the straight line.

From the point A as a centre, and with a radius sufficiently great, describe an arc cutting the line BD in the two points B and D; then mark a point E, equally distant from the points B and

+A

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; hence the line AE is a perpendicular passing through the middle of BD.

PROBLEM.

136. At a point in a given line, to make an angle equal to a given

angle.

Let A be the given point, AB the given line, and IKL the given angle.

K

I A

B

From the vertex K as a centre, with any radius, describe the arc IL, terminating in the two sides of the angle; from the point A as a centre, with a distance 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 one 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; hence they are equal (102.); therefore the angles BAD, IKL, measured by them are equal.

PROBLEM.

137. To divide a given arc, or a given angle, into two equal parts.

First. Let it be required to divide the arc AEB into two equal parts. From the points A and B, as centres, with the same radius, describe two arcs cutting each other in D; through the point D and the centre C, draw CD: it will bisect the arc AB in A the point E.

For the two points C and D are each equally distant from the extremities A and

E

B of the chord AB; hence the line CD bisects the chord at right angles; hence (105.) it bisects the arc AB in the point E.

Secondly. Let it be required to divide the angle ACB into two equal parts. We begin by describing, from the vertex C as a centre, the arc AB; which is then bisected as above. It is plain that the line CD will divide the angle ACB into two equal parts.

138. Scholium. By the same construction, each of the halves AE, EB, may be divided into two equal parts; and