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Arcs and Chords.

Proof. For it has for its measure the half of the semicircumference BED, or a quadrant.

110. Corollary. Every angle BAC (fig. 50) inscribed in a segment greater than a semicircle is an acute angle, for it has for its measure the half of an arc BEC less than a semicircumference.

111. Corollary. Every angle BEC inscribed in a segment less than a semicircle is an obtuse angle; for it has for its measure the half of an arc greater than a semicircumference.

112. Theorem.

In the same circle, or in equal circles, equal arcs are subtended by equal chords. Proof. Let the arc AB (fig. 52) be equal to the arc BC.

Join AC; and, in the triangle ABC, the angles A and C are equal, for they are measured by the halves of the equal arcs BC and AB. The triangle ABC is therefore isosceles, by § 58, and the chords AB and BC are equal.

113. Theorem. Conversely, in the same circle, or in equal circles, equal chords subtend equal arcs.

Proof. Let the chord AB (fig. 52) be equal to the chord BC.

Join AC; and in the isosceles triangle ABC the angles A and C must be equal, by § 55, and also the arcs AB and BC, which are double their measures.

114. Theorem. In the same circle, or in equal circles, if the sum of two arcs be less than a circumference, the greater arc is subtended by the greater chord ; and, conversely, the greater chord is subtended by the greater are.

Perpendicular at the Middle of a Chord.

Proof. a. Let the arc BC (fig. 53) be greater than the arc AB.

Join AC; and the angle BAC, being measured by half the arc BC, is greater than BCA, which is measured by half of AB; and therefore, by §.62, the chord BC is greater than AB.

b. Conversely. Suppose the chord BC AB.

Join AC; and, by § 62, BAC>BCA, and, therefore, the arc BC double the measure of BAC is greater than the arc AB double the measure of BCA.

115. Corollary. If the sum of the two arcs is greater than a circumference, the greater arc is subtended by the less chord, and the less arc by the greater chord.

Proof. Suppose the arc BCNA > BANC (fig. 53). Take ANC from each, and we have the arc BC>BA, and consequently, by the preceding proposition, the chord BC of the less arc BANC is greater than the chord BA of the greater arc BCNA.

116. Theorem. The radius CG (fig. 54), perpendicular to a chord AB, bisects this chord and the arc subtended by it.

Proof. a. The radii CA and CB are equal oblique lines drawn to the chord AB. They are, therefore, by § 38, at equal distances from the perpendicular, or AD = DB.

b. Since the line GC is a perpendicular erected at the middle of the straight line AB, any point of it, as G, is, by § 42, at equal distances from its extremities, that is, the chords AG and GB are equal; and therefore, by § 113, the arcs AG and GB are equal.

Tangent to a Circle.

117. Corollary. The perpendicular erected upon the middle of a chord passes through the centre, and also through the middle of the arc subtended by the chord.

118. Definitions. A secant is a line which meets the circumference of a circle in two points, as AB (fig. 55).

A tangent is a line, which has only one point in common with the circumference, as CD.

The common point M is called the point of contact. Also two circumferences are tangents to each other (figs. 56 and 57), when they have only one point com

mon.

A polygon is said to be circumscribed about a circle, when all its sides are tangents to the circumference ; and in this case the circle is said to be inscribed in the polygon.

119. Theorem. The direction of the tangent is the same as that of the circumference at the point of con

tact.

Proof. Draw through the point M (fig. 55) the secant ME and the tangent MD.

If the secant ME is turned around the point M so as to diminish the angle EMD, the secant ME will approach the tangent MD, and the point E will approach the point M. When ME is turned so far as to pass through the point P next to M, the angle DME will be infinitely small, since P is at an infinitely small distance from M; and the line ME will approach infinitely near the tangent MD, that is, it will, by § 99, coincide with this tangent, which has therefore, by § 11, the same direction with the circumference at M.

Angles formed by Secants and Tangents.

120. Theorem.

The tangent to a circle is perpen

dicular to the radius drawn to the point of contact.

=

Proof. The radius OM : ON (fig. 58) is shorter than any other line, as OP, which can be drawn from the point O to the tangent MP; it is therefore, by § 39, perpendicular to this tangent.

121. The angle BAC (fig. 59), formed by a tangent and a chord, has for its measure half the arc BMA comprehended between its sides.

Proof. a. Draw the diameter AD, and we have

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BAD has, by

But DAC, being a right angle, has for its measure half of a semicircumference, as ABD; also § 107, for its measure half of the arc BD. of BAC is therefore

(ABD-BD) = AMB.

The measure

b. In the same way, it may be shown that BAE has for its measure half the arc BDA.

122. Theorem. The angle BAC, formed by two secants (fig. 60), two tangents (fig. 62), or a tangent and a secant (fig. 61), and which has its vertex without the circumference of the circle, has for its measure half the concave arc BMC intercepted between its sides, minus half the convex arc DNE.

Proof. Join BE; and as BEC is an exterior angle of the triangle ABE, we have, by § 71

whence

BEC ABE + BAC,

=

BAC BEC-ABE.

Angles formed by Chords. Arcs intercepted by Parallels.

But the measure of BEC is half of BMC, and that of ABE is half of DNE; therefore the measure of BAC is

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Scholium. In applying the preceding demonstration to (figs. 61 and 62), the letters B and D must denote the same point; and in (fig. 62) the letters C and E must also denote the same point.

123. Theorem. The angle BAC (fig. 63), formed by two chords, and which has its vertex between the centre and the circumference, has for its measure half the arc BC contained between its sides plus half the arc DE contained between its sides produced.

Proof. Join BE; and, as BAC is an exterior angle of the triangle ABE, we have, by § 71,

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But the measure of BEA is, by § 107, half of BC; and that of ABE is half of DE; therefore the measure of BAC is

} BC +} DE.

124. Theorem. Two parallels AB and DC (figs. 64, 65, 66), intercept upon the circumference equal arcs ᎯᎠ, BC.

Proof. Join BD. The alternate-internal angles ABD and BDC are equal, by § 30; and therefore, the arcs AD and BC, the double of their measures, are equal.

Scholium. In applying this demonstration to figs. (65 and 66), the letters A and B must denote the same point; and in (fig. 66) the letters D and C must also denote the same point.

125. Corollary. The arcs AD and BC (fig. 66) being equal must be semicircumferences, and the chord BC must be a diameter.

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