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

B

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.

Cor. 3. Every angle BAC, inscribed in a segment greater than a semicircle, is an acute angle; for it is measured by 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 half of the arc BAC, greater than a semicircumference.

B

PROPOSITION XIX. THEOREM.

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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, the angle BCD is measured by half the arc BAD; hence the two angles BAD, BCD, ta- D ken together, are measured by the half of the circumference; hence their sum is equal to two right angles.

The angle formed by two chords, which intersect each other, is measured by half the sum of the arcs included between its sides.

Let AB, CD, be two chords intersecting each other at E: then will the angle AEC, or DEB, be measured by half of AC+DB.

PROPOSITION XX. THEOREM.

Draw AF parallel to DC: then will the arc DF be equal to AC (Prop. X.); and the angle FAB equal to the angle DEB (Book I. Prop. XX. Cor. 3.). But F the angle FAB is measured by half the arc FDB (Prop. XVIII.); therefore, DEB is measured by half of FDB; that is, by half of DB+DF, or half of DB+ AC. In the same manner it might be proved that the angle AED is measured by half of AFD+BC.

D

B

Let AB, AC, be two secants: then will the angle BAC be measured by half the difference of the arcs BEČ and DF.

A

Draw DE parallel to AC: then will the arc EC be equal to DF, and the angle BDE equal to the angle BAC. But BDE is measured by half the arc BE; hence, BAC is also measured by half the arc BE; that is, by half the difference of BEC and EC, or half the difference of BEC and DF.

B

The angle formed by two secants, is measured by half the difference of the arcs included between its sides.

PROPOSITION XXI. THEOREM.

C

E

A.

A

E

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

M

Let BE be the tangent, and AC the chord. From A, the point of contact, draw the diameter AD. The angle BAD is a right angle (Prop. IX.), 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.

A

E

It might be shown, by taking the difference between the angles DAE, DAC, that the angle CAE is measured by half the arc AC, included between its sides.

PROBLEM I.

PROBLEMS RELATING TO THE FIRST AND THIRD BOOKS.

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.

C

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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 (Book I. Prop. XVI. Cor.). 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 II.

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.

B 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 c 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, must be in the perpendicular raised from the middle of BC (Book I. Prop. XVI.); and since two points determine a line, AD is that perpendicular.

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

PROBLEM III.

From a given point, without a straight line, to let fall a perpen

dicular on this line.

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

+A

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 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 (Book I. Prop, XVI. Cor.).

PROBLEM IV.

C

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

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

I A

From the vertex K, as a centre, with any radius, describe the arc IL, terminating in the two sides of the angle. From the K point A as a centre, with a disiance AB, equal to Kl, 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 BO, in D; draw AD; and the angle DAB will be equal to the given angle K.

L

PROBLEM V.

OD

J

B

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

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 the point E.

E

For, the two points C and D are each equally distant from the extremities A and B of the chord AB; hence the line CD bisects the chord at right angles (Book I. Prop. XVI. Cor.); hence, it bisects the arc AB in the point E (Prop. VI.). ·

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 AEB; which is then bisected as above. It is plain that the line CD will divide the angle ACB into two equal parts.

Scholium. By the same construction, each of the halves AE, EB, may be divided into two equal parts; and thus, by successive subdivisions, a given angle, or a given arc may be divided into four equal parts, into eight, into sixteen, and so on.

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