Cor. 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 them without the circle, are equal to one another. Thus : 1. 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; 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 the line which meets it, the line which meets, shall touch the circle. B E Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn, of which DCA cuts the circle in the points C, A, and DB meets it in the point B. B F E If the rectangle AD, DC be equal to the square on DB, Construction. Draw the straight line DE, touching the circle ABC, in the point_E (III. 17); find F, the centre of the circle (III. 1), and join FE, FB, FD. Demonstration. Then 1. FED is a right angle (III. 18); and because DE touches the circle ABC, and DCA cuts it, therefore 2. The rectangle AD, DC is equal to the square on DE (III. 36); but the rectangle AD, DC, is, by hypothesis, equal to the square on DB; therefore 3. The square on DE is equal to the square on DB (I. Ax. 1); and the straight line DE equal to the straight line DB. and FE is equal to FB; (I. Def. 15); wherefore DE, EF are equal to DB, BF, each to each; and the base FD is common to the two triangles DEF, DBF; therefore 4. The angle DEF is equal to the angle DBF (I. 8) ; but DEF was shown to be a right angle; therefore also 5. DBF is a right angle (I. 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 6. DB touches the circle ABC. Wherefore, if from a point, &c. Q.E.D. BOOK IV. DEFINITIONS. 1. A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. 2. In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. 3. A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle. 4. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle. о 5. In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure. 6. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. 7. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. PROPOSITION 1.-Problem. In a given circle to place a straight line, equal to a given straight line which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. It is required to place in the circle ABC a straight line equal to D. Construction. Draw BC the diameter of the circle ABC. Then, if BC is equal to D, the thing required is done; for in the circle ABC a straight line BC is placed equal to D. But, if it is not, BC is greater than D (hyp.); make CE equal to D (I. 3); and from the centre C, at the distance CE, describe the circle AEF, and join CA. D E C B Then CA shall be equal to D. Proof. Because C is the centre of the circle AEF, therefore 1. CA is equal to CE (I. Def. 15); but CE is equal to D (constr.); therefore 2. D is equal to CA (Ax. 1). Wherefore in the circle ABC, a straight line CA is placed equal to the given straight line D, which is not greater than the diameter of the circle. Q.E.F. PROPOSITION 2.-Problem. In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle. It is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF C до A Construction. Draw the straight line GAH touching the circle in |