From A and C the extremities of AC the diameter of the circle ABC, let lines AE, EC, AF, FC be drawn to any points E and F in the circumference, and draw the diameter BD perpendicular to AC; join ED, FD; then AE+ EC: AF+ FC :: ED: FD. B/E D H Join AB; and with the centre B and distance BA describe a circle AGC; produce AB, AE, AF to the circumference. Join GH, HI, BE, EF, GI, BF. Then since AG and BD are diameters of the circles, the angles AHG, AIG are equal to DEB, DFB; but BAE, EAF are equal to BDE, EDF, and the angle HIG being = HAG=BDE =BFE, .. the angle HIA = EFD and the triangles GAH, HAI are similar to BDE, EDF, and :. AH : AI :: ED : FD, But (ii. 60.) EH-EC, and FI= FC, (62.) If from any two points in the circumference of a circle there be drawn two straight lines to a point in a tangent to that circle; they will make the greatest angle when drawn to the point of contact. Let A and B be the two points, and CD the tangent at C; join AC, CB; the angle ACB is greater than any other angle ADB formed by lines drawn to any other point D. B Join BE. Then the angles ACB, AEB in the same segment are equal; but ADB is less than the exterior angle AEB, and .. is less than ACB. COR. If two circles touch each other in C, it might be shewn in a similar manner, that the angle formed by two straight lines drawn from A and B to C the point of contact; will be greater than the angle formed by lines drawn from the same points to any point in the exterior circle. (63.) From a given point within a given circle to draw a straight line which shall make with the circumference an angle less than the angle made by any other line drawn from that point. Let P be the given point within the circle ABC. Find the centre, join OP and produce it to the circumference. From P draw PB at right angles to OA; it is the line required. B Join OB, and on it as a diameter describe a circle OPB, which will touch the circle ABC in B. Then OBP is the greatest angle that can be included between lines drawn from O and P to the circumference ABC (ii. 63. Cor.), .. the angle contained by PB and the circumference AB will be the least. (64.) To determine a point in the arc of a quadrant, from which if lines be drawn to the centre and the point of bisection of the radius, they shall contain the greatest possible angle. Let BC be the arch of a quadrant whose centre is A and let the radius AC be bisected in D. On AD describe an equilateral triangle ADE, and produce AE to F; F is the point required. Join FD. Then AF-AC, and AD= AE, but AD is half of AC, and.. AE is half B E of AF, and .. equal to EF; and EA, ED, EF are equal; whence a circle described from the centre E at the distance of any one of them will pass through the extremities of the other two, and touch the arc BC in F, because their centres are in the same straight line; and AFD (ii. 63. Cor.) is greater than any other angle formed by lines drawn from any point in BC to A and D. (65.) If the radius of a circle be a mean proportional to two distances from the centre in the same straight line; the lines drawn from their extremities to any point in the circumference will have the same ratio that the distances of these points from the circumference have. D Let A and B be the two points, such that AO: CO :: CO: BO, and from A and B let lines AD, BD be drawn to any point D in the circumference; these have always the same ratio, viz. AD : BD :: AC: BC. Join OD. Then since OD=OC, OA: OD :: OD : OC, i. e. the sides about the common angle AOD of the triangles AOD, BOD are proportional, and .. the triangles are similar, consequently DA: DB:: OA (OD=) OC. (Eucl. v. 15.) DA: DB:: AC: CB. (66.) Two circles being given in position and magnitude; to draw a straight line cutting them so that the chords in each circle may be equal to a given line, not greater than the diameter of the smaller circle. Let ABC, EFG be the given circles whose centres are O and M. In each place a line AB, EF, equal to the given line; and from the centres draw the perpendiculars OI, MK; and with these distances and centres O and I describe circles which will touch AB, EF in I and K; draw CDGH (ii. 7.) which shall touch these circles in L and N; each of the chords CD and GH will be equal to the given line. Join OL, MN; these lines are perpendicular to CD and GH, and being respectively equal to OI and MK, CD=AB (Eucl. iii. 14.) and GH=EF; but AB and EF are each equal to the given line, .. CD and GH are also each equal to the given line. COR. If the intercepted parts are required to have, a given ratio, take AB and EF in that ratio, and make the same construction as in the proposition. (67.) To determine a point in the arc of a quadrant, through which if a tangent be drawn meeting the sides of the quadrant produced, the intercepted parts may have a given ratio. Let OA, OB be the sides of a quadrant produced; and take M and N two right lines which are in the given ratio, and let OC be a mean proportional between the radius of the quadrant and M, Ε D B and OD a mean proportional between the radius and N. Join CD, and draw the radius OE cutting it at right angles; E is the point required. Through E draw the tangent AEB, which being perpendicular to OE (Eucl. iii. 18.), will be parallel to CD, .. AO: OB :: CO: OD, and since OC and OD are mean proportionals between M and the radius, and N and the radius respectively, M: N in the duplicate ratio of OC: OD, i. e. in the duplicate ratio of AO : OB. But (Eucl. vi. 8. Cor.) AE EB in the duplicate ratio of AO: OB, .. AE: EB:: M: N, i. e. in the given ratio. (68.) If a tangent be drawn to a circle at the extremity of a chord which cuts the diameter at right angles, and from any point in it a perpendicular be let fall; the |