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 (iii) 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 ; subtend equal angles at the centre ; given point to the centre. 3. Enunciate propositions relating to (i) angles in a segment of a circle ; (ii) similar segments of circles. 4. 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 a 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 ; (iv) which are of given radius, and pass through a given point ; (v) which touch a given straight line at a given point ; (vi) which touch each of two parallel straight lines ; (vii) ' which touch each of two intersecting straight lines of unlimited length. 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. H.S.E. Р PROPOSITION 35. THEOREM. 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. H 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 O 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, .:. AB is bisected at G. 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; I. 47. and the sqq. on AG, GF = the sq. on AF ; for the angles at G are rt. angles. .:. 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. But the sq. on AF = the sq. on FD; for AF = 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. = COROLLARY. If through a fixed point within a circle any number of chords are drawn, the rectangles contained by their segments are all equal. NOTE. The following special cases of this proposition deserve notice : (i) when the given chords both pass through the centre : other at right angles : other obliquely. In each of these cases the general proof requires some modification, which may be left as an exercise to the student. · EXERCISES. 1. Two straight lines AB, CD intersect at E, so that the rectangle AE, EB is equal to the rectangle CE, ED; shew that the four points A, B, C, D are concyclic. 2. The rectangle contained by the segments of any chord drawn through a given point within a circle is equal to the square on half the shortest chord which may be drawn through that point. 3. ABC is a triangle right-angled at C; and from C a perpendicular CD is drawn to the hypotenuse : shew that the square on CD is equal to the rectangle AD, DB. 4. ABC is a triangle ; and AP, BQ, the perpendiculars dropped from A and B on the opposite sides, intersect at O: shew that the rectangle AO, OP is equal to the rectangle BO, OQ. 5. Two circles intersect at A and B, and through any point in AB their common chord two chords are drawn, one in each circle ; shew that their four extremities are concyclic. 6. A and B are two points within a circle such that the rectangle contained by the segments of any chord drawn through A is equal to the rectangle contained by the segments of any chord through B: shew that A and B are equidistant from the centre. 7. If through E, a point without a circle, two secants, EAB, ECD are drawn ; shew that the rectangle EA, EB is equal to the rectangle EC, ED. [Proceed as in III. 35, using 11. 6.] 8. Through A, a point of intersection of two circles, two straight lines CAE, DAF are drawn, each passing through a centre and terminated by the circumferences : sħew that the rectangle CA, AE is equal to the rectangle DA, AF. PROPOSITION 36. THEOREM. If from any point without a circle a tangent and a secant are drawn, then the rectangle contained by the whole secant and the part of it without the circle shall be equal to the square on the tangent. Let ABC be a circle; and from D, a point without it, let there be drawn the secant DCA, and the tangent DB. Then the rect. DA, DC shall be equal to the sq. on DB. Construction. Find E, the centre of the O ABC: III. 1. and from E, draw EF perp. to AD. I. 12. Join EB, EC, ED. Proof. Because EF, passing through the centre, is perp. to the chord AC, ... AC is bisected at F. III. 3. And since AC is bisected at F and produced to D, ... the rect. DA, DC with the sq. on FC = the sq. on FD. II. 6. To each of these equals add the sq. on EF: then the rect. DA, DC with the sqq. on EF, FC = the sqq. on EF, FD. But the sqq. on EF, FC = the sq. on EC; for EFC is a rt. angle; = the And the sqq.on EF, FD=the sq. on ED; for EFD is a rt. angle; = the sqq. on EB, BD; for EBD is a rt. angle. III. 18. .. the rect. DA, DC with the sq. on EB=the sqq. on EB, BD. From these equals take the sq. on EB : then the rect. DA, DC=the sq. on DB. Q.E.D. NOTE. This proof may easily be adapted to the case when the secant passes through the centre of the circle. a sq. on EB. a COROLLARY. If from a given point without a circle any number of secants are drawn, the rectangles contained by the whole secants and the parts of them without the circle are all equal ; for each of these rectangles is equal to the square on the tangent drawn from the given point to the circle. For instance, in the adjoining figure, each of the rectangles PB, PA and PD, PC and PF, PE is equal to the square on the tangent PQ : ... the rect. PB, PA - the rect. PD, PC = the rect. PF, PE. = B NOTE. Remembering that the segments into which the chord AB is divided at P, are the lines PA, PB, (see Def., page 139) we are enabled to include the corollaries of Propositions 35 and 36 in a single enunciation. If any number of chords of a circle are drawn through a given point within or without a circle, the rectangles contained by the segments of the chords are equal. EXERCISES. 1. Use this proposition to shew that tangents drawn to a circle from an external point are equal. 2. If two circles intersect, tangents drawn to them from any point in their common chord produced are equal. 3. If two circles intersect at A and B, and PQ is a tangent to both circles ; shew that AB produced bisects PQ. 4. If P is any point on the straight line AB produced, shew that the tangents drawn from P to all circles which pass through A and B are equal. 5. ABC is a triangle right-angled at C, and from any point P in AC, a perpendicular PQ is drawn to the hypotenuse : shew that the rectangle AC, AP is equal to the rectangle AB, AQ. 6. ABC is a triangle right-angled at C, and from C a perpendicular CD is drawn to the hypotenuse : shew that the rect. AB, AD is equal to the square on AC. |