PROP. XXXIII. THEOR. 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 of the line which meets it, the line which meets shall touch the circle. 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, and DB meets it; if the rectangle AD.DC, be equal to the square of DB, DB touches the circle. D Draw the straight line DE touching the circle ABC; find the centre F, and join FE, FB, FD; then FED is a right (18. 3.) angle: and because DE touches the circle ABC, and DCA cuts it, the rectangle AD.DC is equal (32. 3.) to the square of DE; but the rectangle AD.DC is, by hypothesis, equal to the square of DB: therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB: but FE is equal to FB, wherefore DE, EF are equal to DB, BF; and the base FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal (5. 1.) to the angle DBF; and DEF is a right angle, therefore also DBF is a right angle: but FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches (16. 3.) the circle: therefore DB touches the circle ABC. B F E PROBLEMS RELATING TO THE THIRD BOOK. PROP. I. PROBLEM. Having given the circumference or an arc, to find the centre of the given circle. Take any three points A, B, C, in the circumference, or the arc; join AB, BC; and bisect those two lines by the perpendiculars DF, EF: the point F, where these perpendiculars meet, will be the centre sought; for it is equally distant from the points A, B, C. A F E SCHOLIUM. The same construction will serve for describing a circle whose circumference shall pass through three given points, A, B, C. PROP. II. PROB. To draw a straight line from a given point either without or in the circumference, which shall touch a given circle. First, Let A be a given point without the given circle BCD; it is required to draw a straight line from A which shall touch the circle. Find the centre E of the circle, and join AE; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw (Prob. 6. 1.) DF at right angles to EA, join EBF, and draw AB. AB touches the circle BCD. Because E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AEB, FED; therefore the base EF is equal to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles (1. 1.); Therefore the angle EBA is equal to the angle EDF; but EDF is a right angle, wherefore EBA is a right angle; and EB is a line drawn from the centre: but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle (Cor. 16. 3.): Therefore AB touches the circle; and is drawn from the given point A. But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE; DF touches the circle (1. Cor. 16. 3.). SCHOLIUM. When the point A lies without the circle, there will evidently be always two equal tangents passing through the point A. For, by producing the tangent FD till it meets the circumference AG, and joining E and the point of intersection, and also A and the point where this last line will intersect the circumference DC; there will be formed a right angled triangle equal to ABE (Prop. 18. B. I.). PROP. III. PROB. To bisect a given arc, that is, to divide it into two equal parts. Let ADB be the given arc; it is required to bisect it. Join AB, and bisect (Prob. 5. 1.) it in C; from the point C draw CD at right angles to AB, and join AD, DB: the arc ADB is bisected in the point D. Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two sides AC, CD are equal to the two BC, CD; and the angle ACD is equal to the angle BCD, because each of them is a right angle therefore the base AD is equal (1.1.) to the base BD. But equal straight lines cut off equal (27. 3.) arcs, the greater equal to the greater, and the less to the less; and AD, DB are each of them less than a semicircle, because DC passes through the centre; wherefore the arc AD is equal to the arc DB; and therefore the given arc ADB is bisected in D. SCHOLIUM. C B By the same construction, each of the halves AD, DB may be divided into two equal parts; and thus, by successive subdivisions, a given arc may be divided into four, eight, sixteen, &c. equal parts. PROP. IV. PROB. Upon a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle. Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C. H First, let the angle at C be a right angle; bisect (Prob. 5. 1.) AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; the angle AHB being in a semicircle is (29. 3.) equal to the right angle at C. C But if the angle C be not a right angle, at the point A, in the straight line A AB, make (Prob. 9. 1.) the angle BAD equal to the angle C, and from the point A draw (Prob. 6. 1.) AE at right angles to AD; bisect (Prob. 5. 1.) AB in F, and from F draw (Prob. 6. 1.) FG at right angles to AB, and join GB: Then, because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FC; but the angle AFG is also equal to the angle BFG; therefore the base AG is equal (1. 1.) to the base GB; and the circle described from the centre G, at the distance GA, shall pass through the point B; let this be the circle AHB: And because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD (1. Cor. 16. 3.) touches the circle; and because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB (30. 3.); but the angle DAB is equal to the angle C, therefore also the angle C is equal to the angle in the segment AHB: Wherefore, upon the given straight line AB the segment AHB H of a circle is described which contains an angle equal to the given angle at C. PROP. V. PROB. To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the angle D. Draw (Prob. 2. 3.) the straight line EF touching the circle ABC in the point B, and at the point B, in the straight line BF make (Prob. 9. 1.) the angle FBC equal to the angle D; therefore, because the straight line EF To draw a tangent to any point in a circular arc, without finding the centre. From B the given point, take two equal distances BC, CD on the arc; join BD, and draw the chords BC, CD: make the angle CBG=CBD, and the straight line BG will be the tangent required. For the angle CBD=CDB; and therefore the angle GBC is also equal to CDB, an angle in the alternate segment; hence, BG is a tangent at B. G B PROP. VII. PROB. To describe a circle about a given triangle. D Let the given triangle be ABC; it is required to describe a circle about ABC. Bisect (Prob. 5. 1.) AB, AC in the points D, E, and from these points draw DF, EF at right angles (Prob. 6. 1.) to AB, AC; DF, EF produced will meet one another; for, if they do not meet, they are parallel, where fore, AB, AC, which are at right angles to them, are parallel, which is absurd; Let them meet in F, and join FA; also, if the point F be not in BC, join BF, CF: then, because AD is equal to BD, and DF common, |