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 A F B H 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. From F draw FG at rt. angles to AB, cutting AE at G. .. 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 ; BAD the angle in the alt. segment AHB. .. the = But the BAD the C: == .. the angle in the segment AHB = the III. 32. Constr. C. Q.E.F. [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. and Construct a triangle, having given the base, the vertical angle (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 Then arc APB at P: join PX, and produce it to meet the Oce at C. 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 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. PROPOSITION 34. PROBLEM. From a given circle to cut off a scgment which shall contain an angle equal to a given angle. Let ABC be the given circle, and D the given angle. It is required to cut off from the contain an angle equal to D. ABC a segment which shall Construction. Take any point B on the O, and at B draw the tangent EBF. At B, in FB, make the FBC equal to the III. 17. 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, FBC = the angle in the alternate segment BAC. .. the But the the angle in the Hence from the given D; III. 32. Constr. FBC = the segment BAC = the D. 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. QUESTIONS FOR REVISION. 1. Enunciate the propositions from which we infer that a straight line and a circle must either (i) intersect in two points; or (ii) touch at one point; or have no point in common. 2. Give two independent constructions for drawing a tangent to a circle from an external point. Shew that the two tangents so drawn (i) are equal; (ii) subtend equal angles at the centre; (iii) make equal angles with the straight line which joins the given point to the centre. 3. Enunciate propositions relating to 4. (i) angles in a segment of a circle ; What are conjugate arcs of a circle? The angles in conjugate segments of a circle are supplementary. How does Euclid enunciate this theorem? State and prove its converse. 5. Explain what is meant by a reflex angle. What simplifications may be made in the proofs of Third Book Propositions if reflex angles are admitted? 6. If the circumference of a circle is divided into six equal arcs, shew that the chords joining successive points of division are all equal to the radius of the circle. 7. Find the locus of the centres of all circles (i) which pass through two given points; (ii) which touch a given circle at a given point; (iii) which are of given radius, and touch a given circle ; point; (v) which touch a given straight line at a given point; (vii) which touch each of two intersecting straight lines of 8. If a system of triangles stand on the same base and on the same side of it, and have equal vertical angles, shew that the locus of their vertices is the arc of a circle. Prove this theorem, having first enunciated the proposition of which it is the converse. If two chords of a circle cut one another, the rectangle contained by the segments of one shall be equal to the rectangle contained by the segments of the other. Let AB, CD, two chords of the O ACBD, cut one another at E. Then shall the rect. AE, EB = the rect. CE, ED. Construction. Find F, the centre of the ACB; III. 1. From F draw FG, FH perp. respectively to AB, CD. I. 12. Join FA, FE, FD. Proof. Because FG is drawn from the centre F perp. to AB, III. 3. For a similar reason CD is bisected at H. Again, because AB is divided equally at G, and unequally at E, ... the rect. AE, EB with the sq. on EG= the sq. on AG. II. 5. To each of these equals add the sq. on GF; then the rect. AE, EB with the sqq. on EG, GF = the sum of the sqq. on AG, GF. But the sqq. on EG, GF = the sq. on FE; and the sqq. on AG, GF = the sq. on AF ; for the angles at G are rt. angles. I. 47. the rect. AE, EB with the sq. on FE = the sq. on AF. Similarly it may be shewn that the rect. CE, ED with the sq. on FE the sq. on FD. .. the rect. AE, EB with the sq. on FE = the rect. CE, ED with the sq. on FE. From these equals take the sq. on FE: then the rect. AE, EB = the rect. CE, ED. Q.E.D. |