On the radius AB of the quadrant AGB let the semicircle AEB be described, and at A draw the tangent AD. From B draw any line BECD meeting the tangent in D, and the circumferences E B in C, E; from E let fall the perpendicular EF; then BD, BC, BE, BF are in continued proportion. Since FE is perpendicular to BA, it is parallel to AD, .. BF: BE:: (BA=) BC : BD, But (Eucl. vi. 8.) BF: BE :: BE: (BA=) BC, ..(Eucl. v. 15.) also BE: BC :: BC: BD, and BF: BE:: BE: BC:: BC BD. (59.) If the chord of a quadrant be made the diameter of a semicircle, and from its extremities two straight lines be drawn to any point in the circumference of the semicircle; the segment of the greater line intercepted between the two circumferences shall be equal to the less of the two lines. Let O be the centre of the quadrant ADB, join AB and on it let a semicircle ACB be described; from any point C in which let lines CA, CB be drawn to A and B, of which CB is the greater; CD=CA. D B Join AD, and complete the circle ABE; take any point E and join EA, EB. Since ADBE is a quadrilateral figure inscribed in a circle, the angles AEB, ADB are equal to two right angles, and .. equal to ADB, ADC; whence AEB=ADC; but AEB is half of AOB which is a right angle, .. ADC is half a right angle, and ACD being a right angle (Eucl. ii. 31.), CAD is half a right angle, and.. equal to CDA, consequently CA CD. (60.) If two circles cut each other so that the circumference of one passes through the centre of the other, and from either point of intersection a straight line be drawn cutting both circumferences; the part intercepted between the two circumferences will be equal to the chord drawn from the other point of intersection to the point where it meets the inner circumference. Through the centre of the circle ABC, let the circle AOB be described, cutting ABC in A and B. If any line AED be drawn from A, and BE joined; DE will be equal to EB. D B Then Draw the diameter AOC; join BC, BD. since the angle AOB is equal to AEB, .. the angle COB is equal to DEB. Also the angles OCB, EDB, being in the same segment, are equal to one another, .. the triangles OCB, EDB are equiangular, and .. since OB=OC, the angle OCB is equal to the angle OBC, whence EDB = EBD, and .. ED=EB. (61.) If from each extremity of the diameter of a circle lines be drawn to any two points in the circumference; the sums of the lines so drawn to each point will have to one another the same ratio that the lines have, which join those points and the opposite extremity of a diameter perpendicular to the former. From A and C the extremities of AC the diameter of the circle ABC, let lînes 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 F D C 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 exteriorcircle. (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. 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 cir- 62 cumference 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. 62 Let BC be the arch of a quadrant whose centre is A Join FD. Then AF-AC, and AD= B 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. Let A and B be the two points, such that 40: CO :: ÇO: BO, and from A and B let lines AD, BD be drawn to any point D in the circum D D ference; these have always the same ratio, viz. AD : BD :: AC : BC. Join OD. Then since OD=OC, OA : OD :: OD : OB 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. |