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If a straight line touches a circle, and from the point of contact a chord is drawn, the angles which this chord makes with the tangent shall be equal to the angles in the alternate segments of the circle.

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Let EF touch the given o ABC at B, and let BD be a chord drawn from B, the point of contact. Then shall

(i) the 2 DBF the angle in the alternate segment BAD : (ii) the 2 DBE = the angle in the alternate segment BCD.

I. 11.

Construction. From B draw BA perp. to EF.

Take any point C in the arc BD;

and join AD, DC, CB.

..

(i) Proof. Because BA is drawn perp. to the tangent EF, at its point of contact B,

BA passes through the centre of the circle: III. 19. ... the 2 ADB, being in a semicircle, is a rt. angle: III. 31. ... in the A ABD, the other <* ABD, BAD together =a rt. angle;

I. 32. that is, the <* ABD, BAD together= the 2 ABF.

From these equals take the common – ABD ; ::. the < DBF = the į BAD, which is in the alternate seg(ii) Because ABCD is a quadrilateral inscribed in a circle, .. the opp. 2* BCD, BAD together = two rt, angles : III. 22.

ment.

but the 4* DBE, DBF together = two rt. angles ; I. 13. .. the < DBE, DBF together= the < BCD, BAD;

and of these the 2 DBF = the 2 BAD; Proved. .. the _ DBE = the _ BCD, which is in the alternate segment.

Q.E.D.

EXERCISES.

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1. State and prove the converse of Proposition 32.

2. Use this proposition to shew that the tangents drawn to a circle from an external point are equal.

3. If two circles touch one another, any straight line drawn through the point of contact cuts off similar segments.

Prove this for (i) internal, (ii) external contact.

4. If two circles touch one another, and from A, the point of contact, two chords APQ, AXY are drawn : then PX and QY are parallel.

Prove this for (i) internal, (ii) external contact.

5. Two circles intersect at the points A, B: and one of them passes through O, the centre of the other : prove that OA bisects the angle between the common chord and the tangent to the first circle at A.

6. Two circles intersect at A and B; and through P, any point on the circumference of one of them, straight lines PAC, PBD are drawn to cut the other circle at C and D: shew that CD is parallel to the tangent at P.

7. If from the point of contact of a tangent to a circle, a chord is drawn, the perpendiculars dropped on the tangent and chord from the middle point of either arc cut off by the chord are equal.

PROPOSITION 33.

PROBLEM. On a given straight line to describe a segment of a circle which shall contain an angle equal to a given angle. H

H

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I. 10.

Let AB be the given st. line, and C the given angle. It is required to describe on AB a segment of a circle which shall contain an angle equal to C. Construction. At A in BA; make the 2 BAD equal to the L C. I. 23. From A draw AE at rt. angles to AD.

1. 11. Bisect AB at F. From F draw FG at rt. angles to AB, cutting AE at G.

Join GB.
Then in the AS AFG, BFG,
AF = BF,

Constr.
Because

and FG is common, and the 2 AFG = the 2 BFG, being rt. angles ;

GA = GB: .. the circle described with centre G, and radius GA, will pass through B.

Describe this circle, and call it ABH.
Then the segment AHB shall contain an angle equal to C.

Proof. Because AD is drawn at rt. angles to the radius GA from its extremity A, .. AD is a tangent to the circle ;

III, 16. and from A, its point of contact, a chord AB is drawn ; ... the 2 BAD = the angle in the alt. segment AHB. But the BAD == the LC:

Constr. .. the angle in the segment AHB= the < C.

.: AHB is the segment required. Q.E.F.

I, 4.

III. 32. NOTE. In the particular case when the given angle C is a rt. angle, the segment required will be the semicircle described on the given st. line AB; for the angle in a semicircle is a rt. angle.

III. 31.

A

B

EXERCISES.

[The following exercises depend on the corollary to the Converse of Proposition 21 given on page 201, namely

The locus of the vertices of triangles which stand on the same base and have a given vertical angle, is the arc of the segment standing on this base, and containing an angle equal to the given angle.

Exercises 1 and 2 afford good illustrations of the method of finding required points by the Intersection of Loci. See page 125.]

1. Describe a triangle on a given base, having a given vertical angle, and having its vertex on a given straight line.

2. Construct a triangle, having given the buse, the vertical angle and (i) one other side.

(ii) the altitude.
(iii) the length of the median which bisects the base.
(iv) the point at which the perpendicular from the vertex meets

the base. 3. Construct a triangle having given the base, the vertical angle, and the point at which the base is cut by the bisector of the vertical angle.

[Let AB be the base, X the given point in it, and K the given angle. On AB describe a segment of a circle containing an angle equal to K; complete the Oce by drawing the arc APB. Bisect the arc APB at P: join PX, and produce it to meet the Oce at C. Then ABC shall be the required triangle.)

4. Construct a triangle having given the base, the vertical angle, and the sum of the remaining sides.

(Let AB be the given base, K the given angle, and H the given line equal to the sum of the sides. On AB describe a segment containing an angle equal to K, also another segment containing an angle equal to half the 2 K. From centre A, with radius H, describe a circle cutting the arc of the last drawn segment at X and Y. Join AX (or AY) cutting the arc of the first segment at C. Then ABC shall be the required triangle.]

5. Construct a triangle having given the base, the vertical angle, and the difference of the remaining sides.

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From a given circle to cut off a scgment which shall contain an angle equal to a given angle.

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Let ABC be the given circle, and D the given angle. It is required to cut off from the O ABC a segment which shall

contain an angle equal to D.

Construction. Take any point B on the Oce,
and at B draw the tangent EBF.

III. 17. At B, in FB, make the 2 FBC equal to the 2 D. I. 23. Then the segment BAC shall contain an angle equal to D.

Proof. Because EF is a tangent to the circle, and from B, its point of contact, a chord BC is drawn, .. the 2 FBC= the angle in the alternate segment BAC.

III. 32. But the L FBC=the _D;

Constr. the angle in the segment BAC = the 2 D. Hence from the given O ABC a segment BAC has been cut off, containing an angle equal to D.

Q.E.F.

EXERCISES.

1. The chord of a given segment of a circle is produced to a fixed point: on this straight line so produced draw a segment of a circle similar to the given segment.

2. Through a given point without a circle draw a straight line that will cut off a segment capable of containing an angle equal to a given angle.

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