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342. The perpendicular from the center of a circle to a secant bisects the chord; and, if a line through the center bisect a chord not passing through the center, it cuts it at right angles.

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PROOF. For any chord, there is only one perpendicular from the center, only one line through the mid-point and center, only one perpendicular bisector; and these, by 331, are identical.

(331. The perpendicular bisector of any chord passes through the center.)

343. Every diameter is an Axis of Symmetry.

For if we fold over along a diameter, every point on the part of the circle turned over must fall on some point on the other part, since its sect from the center, which remains fixed, is a radius.

INVERSE. Every line which is an axis of symmetry of a circle contains a diameter.

For, if not, there would be a point symmetrical to the center, and this, too, would again be a center: the circle would thus have two centers.

(330. A circle has only one center.)

Each circle has therefore, besides its center of symmetry, an infinite number of axes of symmetry.

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HYPOTHESIS. Let A and B be any two points in O ABC.

CONCLUSION. Every point on chord AB between A and B is within the O ABC.

PROOF. Take any point, D, in chord AB.
By 332, find 0, the center of the circle.
Join 0A, OD, OB.

Then OB makes a greater sect than OD from the foot of the perpendicular from on the line AB,

(342. The perpendicular from center on line AB bisects chord AB.)

:. OB > OD, (154. The oblique which makes the greater sect from the foot of the perpendicular

is the greater.)

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(320. A point is within the circle if its sect from the center is less than the radius.)

345. COROLLARY. If a line has a point within a circle, it is a secant, for the radius is greater than the perpendicular from the center to this line: so there will be two sects from the center to the line, each equal to the radius; that is, the line will pass through two points on the circle.

Thus, again, the circle is a closed curve.


346. In a circle, two chords which are not both diameters do not mutually bisect each other.

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HYPOTHESIS. Let the chords AB, CD, which do not both pass through the center, cut one another in the point F, in the O ACBD.

CONCLUSION. AB and CD do not mutually bisect each other.

PROOF. If one of them pass through the center, it is not bisected by the other, which does not pass through the center.

If neither pass through the center, find the center O, and join OF. If F is the bisection point of one of the chords, as AB, then

X OFB = rt. 4, (342. If a line through the center bisect a chord not passing through the center, it

cuts it at right angles.)

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EXERCISES. 69. What is the locus of mid points of parallel chords?

70. Prove by symmetry that the diameter perpendicular to a chord bisects that chord, bisects the two arcs into which this chord divides the circle, and bisects the angles at the center subtended by these arcs.


347. If from any point not the center, sects be drawn to the different points of a circle, the greatest is that which meets the circle after passing through the center; the least is part of the same line.

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HYPOTHESIS. From any point A, sects are drawn to a O DBHK, whose center is C.

Then, because CB = CD,

... AB = AC + CB = AC + CD > AD.
(156. Any two sides of a triangle are greater than the third.)

PROOF. When A is within the circle,

HC = KC <KA + AC, by 156.
Taking away AC from both sides,

.. AH < AK. When A is on the circle, AH is a point. When A is without the circle,

AC = AH + HC < AK + KC, by 156. Taking away HC = KC,

AH < AK. 348. COROLLARY. The diameter of a circle is greater than any other chord.



349. If from any point three sects drawn to a circle are equal, that point is the center.

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HYPOTHESIS. From the point 0 to O ABC, let OA = OB = OC.
CONCLUSION. O is center of O ABC.
PROOF. Join AB, and BC.

O is on the perpendicular bisector of chord AB, and also on that of chord BC, (183. The locus of a point to which sects from two given points are equal the

perpendicular bisector of the sect joining them.)

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(332. The center is the intersection of perpendicular bisectors of two non-parallel


350. CONTRANOMINAL OF 349. From any point not the center, there cannot be drawn more than two equal sects to a circle.

351. COROLLARY I. If two circles have three points in common, they coincide.

Because from the center of one circle three equal sects can be drawn to points on the other circle, :: they are concentric, and they have a point in common,

.. they coincide.
(340. Two concentric circles with a point in common coincide.)

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