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CONCLUSIONS. * DCB

* DCB = 1 X DOC.

* DCA = { explement DOC.

PROOF. At C erect chord CF I AB.

CF is a diameter of the circle.

(393. The perpendicular to a tangent from the point of contact passes through the

center of the circle.)

Join OD. Then

rt. X OCA = 1 st. 4 at 0, also

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(375. The angle at the center is double the inscribed angle on the same arc.)

Therefore, adding

* DCA = reflex DOC, therefore the supplement of X, DCA, which is XDCB, is half the explement of reflex X DOC, which is X DOC.

398. INVERSE. If the angle at the center standing on the arc intercepted by a chord equals twice the angle made by that chord and a line from its extremity on the same side as the arc, this line is a tangent.

PROOF. There is but one line which will make this angle; and we already know, from 397, that a tangent makes it.

EXERCISES. 73. The chord which joins the points of contact of parallel tangents to a circle is a diameter.

74. How may 397 be considered as a special case of 375 ?

75. A parallelogram inscribed in a circle must be a rectangle.

76. If a series of circles touch a given line at a given point, where will their centers all lie?

77. The angle of two tangents is double that of the chord of contact and the diameter through either point of contact.

THEOREM XXIV.

399. The angle formed by a tangent and a secant is half the angle at the center standing on the difference of the intercepted

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HYPOTHESIS. Of two lines from D, one cuts at B and C theo whose center is 0, the other is tangent to the same o at A.

CONCLUSION. 2 * D = difference between * AOC and 4 AOB. PROOF. Join AB.

* ABC = %D + DAB,

(173. The exterior angle of a triangle is equal to the two interior opposite angles.)

.. AOC = 2% D + 4 AOB,

(375. An angle at the center is double the inscribed angle on the same arc.) and (397. An angle formed by a tangent and chord is half the angle at the center on

the intercepted arc.)

i. twice D is the difference between 4 AOC and X AOB.

400. COROLLARY. The angle formed by two tangents is half the angle at the center standing on the difference of the intercepted arcs.

PROBLEM II.

401. From a given point without a circle to draw a tangent to the circle

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GIVEN, a O with center 0, and a point P outside it.
REQUIRED, to draw through Pa tangent to the circle.
CONSTRUCTION. Join OP. Bisect OP in C.

With center C and radius CP describe a circle cutting the given circle in F and G.

Join PF and PG.

These are tangents to O FGB.
PROOF. Join OF.

* OFP is a rt. *,

(383. The angle in a semicircle is a right angle.)

.. PF is tangent at F to O BFG.

(389. A line perpendicular to a radius at its extremity is a tangent to the circle.)

402. COROLLARY. Two tangents drawn to a circle from the same external point are equal, and make equal angles with the line joining that point to the center.

V. Two Circles.

THEOREM XXV.

403. The line joining the centers of two circles which meet in two points is identical with the perpendicular bisector of the common chord.

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PROOF. For the perpendicular bisector of the common chord must pass through the centers of the two circles.

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

404. COROLLARY.

Since any common chord is bisected by the line joining the centers, therefore if the two circles meet at a point on the line of centers, there is no common chord, and these circles have no second point in common.

405. Two circles which meet in one point only are said to touch each other, or to be tangent to one another, and the point at which they meet is called their point of contact.

406. By 343,

Two circles, not concentric, have always one, and only one, common axis of symmetry; namely, their line of centers.

For this is the only line which contains a diameter of each.

THEOREM XXVI.

OBVERSE OF 404.

407. If two circles have one common point not on the line through their centers, they have also another common point.

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HYPOTHESIS.

with center and with center C, having a common point B not on OC.

CONCLUSION. They have another common point.

PROOF. Join OC, and from B drop a line perpendicular to OC at D, and prolong it, making DF = BD.

F is the second common point. For A ODB=A OFD, and A CBDSA CFD; (124. Triangles having two sides and the included angle equal in each are con

gruent.)

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(321. A point is on the circle if its distance from the center is equal to the radius.)

408. CONTRANOMINAL OF 407. If two circles touch one another, the line through their centers passes through the point of contact.

409. COROLLARY. Two circles which touch one another have a common tangent at their point of contact; namely, the perpendicular through that point to the line joining their centers.

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