of them without the circle shall be equal to one another. Let A be any point without the circle BEF; and from A let AEB, AFC be drawn cutting the circle in E, B and F, c respectively. Then the rectangle BA, AE shall be equal to the rectangle CA, AF. From A draw (iii. 17) AD touching the circle at D. Then by the prop1 each of the rectangles BA, AE and CF, FA is equal to the square of the touching line AD; and things that are D B E F equal to the same thing are equal to one another (AX. 1): therefore the rectangle BA, AE is equal to the rectangle CA, AF. Which was to be proved. PROP. XXXVII. THEOR. If from any point without a circle two straight lines be drawn, one of which cuts the circle and the other meets it, such that the rectangle contained by the whole line cutting the circle and the part of it without the circle is equal to the square of the meeting line: then the meeting line shall touch the circle. Let D be any point without the circle ABC, and from D let the straight lines DCA, DB be drawn, one of which DCA cuts the circle in c and A, and the other DB meets the circle in B, so that the rectangle contained by the whole of the cutting line DA and the part of it DC without the circle is equal to the square of the meeting line DB. Then DB shall touch the circle at B. Find (iii. 1) the centre F of the circle ABC; join DF; and from D draw (iii. 17) the straight line DE touching the circle in E on the side of DF opposite to DB. FB, FE. Join Because DE touches the circle by const" and FE is drawn from the centre F to the point of contact E, the angle FED is a right angle (iii. 18). And therefore the square of DE is equal to the square of ᎠᏴ, and hence DE is equal to DB. Also FB is equal to FE by the def" of a circle, and FD is common to the two triangles DEF, DBF therefore these two triangles have the three sides DE, EF, FD respectively equal to the three sides DB, BF, FD. Therefore they are equal in every respect (i. 8); and hence the angle DEF is equal to the angle DBF. But DEF is a right angle: therefore also DBF is a right angle. Hence BD is drawn perpendicular to the radius BF of the circle from its extremity B: therefore BD touches (iii. 16, Cor.) the circle ABC at B. was to be proved. Which N THE ELEMENTS OF EUCLID. BOOK IV. DEFINITIONS. I. A POLYGON is said to be inscribed in another polygon of the same number of sides, when each of the angular points of the former is in a side of the latter. II. A polygon is said to be circumscribed about another polygon of the same number of sides, when each of the sides of the former passes through an angular point of the latter. III. A polygon is said to be inscribed in a circle, when all its angular points are in the circumference of the circle. IV. A polygon is said to be circumscribed about a circle, when each of its sides touches the circle. V. A circle is said to be inscribed in a polygon, when it is touched by each of the sides of the polygon. VI. A circle is said to be circumscribed about a polygon, when the circumference of the circle passes through all the angular points of the polygon. VII. A straight line is said to be placed in a circle, when its extremities are in the circumference of the circle. PROPOSITIONS. PROP. I. PROB. 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 which is not greater than the diameter of the circle. It is required to place in the circle ABC a straight line equal to D. Find (iii. 1) the centre of the circle ABC, and through it draw any diameter BC: then if Bс be equal to D, the thing required is already done. But if not: BC must be greater than D by hyps; from CB cut off (i. 3) CE equal to D, and A F D E B with centre c and radius CE describe the circle FEA, cutting the circle ABC in a. straight line required. Join CA. Then CA shall be the By the def" of a circle ca is equal to CE; and D is equal to CE by const"; and things that are equal to the same thing are equal to one another (Ax. 1): therefore CA is equal to D, and its extremities are points in the circumference of the circle ABC. Hence in the given circle ABC has been placed (iv. Def. 7) a straight line ac equal to the given straight line D. Which was to be done." PROP. II. PROB. In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given tri |