and radius OC, describe a circle CFG, cutting AB in F and G; join CF, CG; they are the lines required. For (Eucl. vi. C.) the rectangle CF, CG is equal to the rectangle contained by 2 CO and CE, i. e. to the given rectangle. And since AD = DB, and FD=DG, .. AF GB. = (13.) To draw a straight line which shall touch a given circle, and make with a given line, an angle equal to a given angle. D E Let AB be the given line, and O the centre of the given circle. From any point A in the given line, draw AC making with it an angle equal to the given angle; from O draw OD perpendicular to AC, and through the point E where it meets the circle, draw EF parallel to DA; EF is the line required. A F B For being parallel to AC it is perpendicular to OD, and a tangent to the circle; and the angle EFB = DAB the given angle. = (14.) Through a given point to draw a line terminating in two lines given in position, so that the rectangle contained by the two parts may be equal to a given rectangle. Let AB, CD be the lines given in position, E the given point; from E draw EF perpendicular to AB, and Join GH. The angle GEH is equal to AEF, and the angles GHE, AFE are right angles, .. the triangles GEH, AEF are equiangular, and EH EG: EF : EA, whence the rectangle AE, EH is equal to the rectangle EG, EF, i. e. to the given rectangle. (15.) From a given point to draw a line cutting two given parallel lines, so that the difference of its segments may be equal to a given line. Let AB, CD be the given parallels, and P the given point. From P draw any line PB, meeting the given. lines in B and E. Make EF = EP, and draw FG G parallel to AB. With any point O as centre, and radius equal to the given line, describe a circle cutting GF in H. Join OH, and draw PGA parallel to it. PGA will be the line required. Since PE is equal to EF, .. (Eucl. vi. 2.) PI=IG; and AG is equal to the difference of AI and IP, the segments of PA; and AG=OH= the given line. (16.) From a given point without a circle, to draw a straight line cutting the circle, so that the rectangle contained by the part of it without and the part within the circle shall be equal to a given square. B Let A be the given point, and BCD the given circle. From A draw AB touching the circle; and on it as a diameter describe a semicircle, cutting the given circle in C. Join AC, and produce it to D. Now if the side of the given square be equal to BC, the problem is possible. For the rectangle AC, AD is equal to the square of AB, i. e. to the squares of AC and BC; take away from each the square of AC, .. the rectangle AC, CD is equal to the square of BC. (17.) From a given point in the circumference of a semicircle, to draw a straight line meeting the diameter, so that the difference between the squares of this line and a perpendicular to the diameter from the point of intersection may be equal to a given rectangle. DB Let A be the given point in the circumference of the semicircle; from it draw AD perpendicular to the diameter. Take the centre, and divide DO in B, so that the rectangle contained by 2OB, BD may be equal to the given rectangle. Join AB; and draw BC perpendicular to DB. AB, BC are the lines required. For (Eucl. ii. 12.) the square of AB together with twice the rectangle OB, BD is equal to the difference of the squares of OA and OB, i. e. to the square of BC; ..the difference between the squares of AB and BC is equal to twice the rectangle OB, BD, i. e. to the given rectangle. (18.) From a given point to draw two lines to a third given in position, so that the rectangle contained by those lines may be equal to a given rectangle, and the difference of the angles which they make with that part of the third which is intercepted between them may be equal to a given angle. B H E Let A be the given point, and BC the line given in position. From A draw AD perpendicular to BC; make the angle DAE equal to the given angle; and produce AE, till the rectangle DA, AE, is equal to the given rectangle. On AE as a diameter describe the circle AFG, cutting BC in F and G. Join AF, AG; they are the lines required. Draw GH perpendicular to the diameter AE; then the arc HA is equal to the arc AG, and the angle AGH to AFG;. the angle HGF is equal to the difference of the angles AGF, AFG. Now the right-angled triangles AIK, KDG have the angles at K equal, .. the angle KAI=KGD; but KAI was made equal to the given angle; .. the difference of the angles AFG, AGF is equal to the given angle. And (Eucl. vi. C.) the rectangle AF, AG is equal to the rectangle DA, AE, i. e. to the given rectangle. (19.) Two points being given without a given circle; to determine a point in the circumference, from which lines drawn to the two given points shall contain the greatest possible angle. Let A and B be the given points, and EDF the given circle whose centre is 0. Describe a circle through A, B, O. Join EF, BA, and produce them to meet in G. From G draw GD touching the given circle in D. Through D, A, B describe another circle; then since the square of GD is equal to the rectangle EG, GF, i. e. to the rectangle AG, GB, .. GD touches the circle ABD. Join AD, DB. D is the point required, as is evident from (ii. 2.) (20.) From the bisection of a given arc of a circle to draw a straight line such that the part of it intercepted between the chord of that and the opposite circumference shall be equal to a given straight line. Let DAE be the given arc of the H circle ABC, bisected in A; AFC the diameter, and HI the given straight line. Produce HI to K, so that the rectangle HK, KI may be equal to the rectangle D B E FA, AC. From A place in the circle AB=HK; AB is the line required. Join BC; then the angle AFG being a right angle is equal to the angle ABC, and the angle at A is common,.. the triangles AGF, ABC are equiangular, |