CASE III.-Let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in the point E, but not at right angles. CONSTRUCTION.-Bisect BD in F, then F is the centre of the circle. Join AF, and from F draw FG dicular to AC (I. 12). perpen PROOF. Then AG is equal to GC (III. 3). Bisect BD in F the centre. Draw FG F E G B AC. at right angles to ... AG = GC. Therefore the rectangle AE, EC, together with the square Now, on EG, is equal to the square on AG (II. 5). To each of these equals add the square on GF; But the squares on EG, GF are equal to the square on EF; AE EC + = AG2. +EF= Therefore the rectangle AE, EC, together with the square. AE-EC on EF, is equal to the square on AF; that is, the square on FB. But the square on FB is equal to the rectangle BE, ED, together with the square on EF (II. 5); Therefore the rectangle AE, EC, together with the square on EF, is equal to the rectangle BE, ED, together with the square on EF. Take away the common square on EF; AF2 = FB2 = BE ED + EF2. And the remaining rectangle AE, EC is equal to the .. AE EC remaining rectangle BE, ED (Ax. 3). CASE IV. Let neither of the straight lines AC, BD pass through the centre. CONSTRUCTION.-Take the centre F (III. 1), and through E, the intersection of the lines AC, BD, draw the diameter A GEFH. PROOF. Because the rectangle GE, EH is equal, as has been shown, to the H B rectangle AE, EC, and also to the rectangle BE, ED; = BE⚫ED. Therefore the rectangle AE, EC is equal to the rectangle ..AE EC BE, ED (Ax. 1). Therefore, if two straight lines, &c. Q.E.D. = BE ED. Proposition 36-Theorem, If from a point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square on the line which touches it. Let D be any point without the circle ABC, and let DCA, DB be two straight lines drawn from it, of which DCA cuts the circle, and DB touches it. AD DC + B ... AD DC + EB2 = BD2+EB2. ... AD DC = BD2. The rectangle AD, DC shall be equal to the square on DB, CASE I. Let DCA pass through the centre E, and join EB. D D B A PROOF. Then EBD is a right angle (III. 18). And because the straight line AC is bisected in E, and produced to D, the rectangle AD, DC, together with the square on EC, is equal to the square on ED (II. 6). But EC is equal to EB; Therefore the rectangle AD, DC, together with the square on EB, is equal to the square on ED. But the square on ED is equal to the squares on EB, BD, because EBD is a right angle (I. 47); Therefore the rectangle AD, DC, together with the square on EB, is equal to the squares on EB, BD. Take away the common square on EB; Then the remaining rectangle AD, DC is equal to the square on DB (Ax. 3). CASE II.-Let DCA not pass through the centre of the circle ABC. (III. 1), and draw EF perpendicular to AC (I. 12), and join EB, EC, ED. PROOF. Because the straight line EF, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it also bisects it (III. 3); Therefore AF is equal to FC. .AF= FC. ... + FC2= FD2. And because the straight line AC is bisected in F and produced to D, the rectangle AD, DC, together with the ADDC square on FC, is equal to the square on FD (II. 6). To each of these equals add the square on FE; Therefore the rectangle AD, DC, together with the squares on CF, FE, is equal to the squares on DF, FE (Ax. 2). But the squares on CF, FE are equal to the square on CE, because CFE is a right angle (I. 47); And the squares on DF, FE are equal to the square on DE; Therefore the rectangle AD, DC, together with the square on CE, is equal to the square on DE, But CE is equal to BE; Therefore the rectangle AD, DC, together with the square on BE, is equal to the square on DE. But the square on DE is equal to the squares on DB, BE, because EBD is a right angle (I. 47); ... AD.DC DE2. Therefore the rectangle AD, DC, together with the square. AD DU + BE2 = DB2+BE2 on BE, is equal to the squares on DB, BE. Take away the common square on BE; Then the remaining rectangle AD, DC is equal to the .. AD DC square on DB (Ax. 3). Therefore, if from any point, &c. Q. E.D. COROLLARY.-If from any point without a circle there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of D them without the circle, are equal to one another; namely, the rectangle BA, AE is equal to the rectangle CA, AF; for each of them is equal to the square on the straight line AD, which touches the circle. Proposition 37.-Theorem. If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on = DB2. Draw DE touching the circle. Then the line which meets the circle, the line which meets the circle shall touch it. Let any point D be taken without the circle ABC, and from it let two straight lines, DCA, DB, be drawn, of which DCA cuts the circle, and DB meets it; and let the rectangle AD, DC be equal to the square on DB. Then DB shall touch the circle. CONSTRUCTION.-Draw the straight line DE, touching the circle ABC (III. 17); Find F the centre (III. 1` and join FB, FD, FE. PROOF. Then FED is a right angle (III. 18). And because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE (III. 36). But the rectangle AD, DC is equal to square on DB (Hyp.); the Therefore the square on DE is equal to the square on DB (Ax. 1); Therefore the straight line DE is equal to the straight line DE=DB. DB. And triangles DBF and DEF are equal in every respect. .. DBF is a right angle; and therefore ᎠᏴ touches the circle. And EF is equal to BF (I. Def. 15); Therefore the two sides DE, EF are equal to the two sides DB, BF, each to each; And the base DF is common to the two triangles DEF, Therefore the angle DEF is equal to the angle DBF (I. 8). Therefore also DBF is a right angle (Ax. 1). And BF, if produced, is a diameter; and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches the circle (III. 16, Cor.); Therefore DB touches the circle ABC. EXERCISES ON BOOK III. PROP. 1-15. 1. Two straight lines intersect. Describe a circle passing through the point of intersection and two other points, one in each straight line. 2. If two circles cut each other, any two parallel straight lines drawn through the points of section to cut the circumferences are equal. 3. Show that the centre of a circle may be found by drawing perpendiculars from the middle points of any two chords. 4. Through a given point, which is not the centre, draw the least line to meet the circumference of a given circle, whether the given point be within or without the circle. 5. The sum of the squares of any two chords in a circle, together with four times the sum of the squares of the perpendiculars on them from the centre, is equal to twice the square of the diameter. 6. With a given radius, describe a circle passing through the centre of a given circle and a point in its circumference. 7. If two chords of a circle are given in magnitude and position, describe the circle. 8. Describe a circle which shall touch a given circle in a given point, and shall also touch another given circle. 9. If, from any point in the diameter of a circle, straight lines be drawn to the extremities of a parallel chord, the squares of these lines are together equal to the squares of the segments into which the diameter is divided. 10. If two circles touch each other externally, and parallel diameters be drawn, the straight line joining extremities of these diameters will pass through the point of contact. 11. Draw three circles of given radii touching each other. 12. If a circle of constant radius touch a given circle, it will always touch the same concentric circle. 13. If a chord of constant length be inscribed in a circle, it will always touch the same concentric circle. 14. The locus of the middle points of chords parallel to a given straight line is a line drawn through the centre perpendicular to the parallel chords. PROP. 16-30. 15. Show that the two tangents from an external point are equal in length. 16. Draw a tangent to a given circle, making a given angle with a given straight line. 17. If a polygon having an even number of sides be inscribed in a circle, the sums of the alternate angles are equal. |