At C and D the extremities of the chord CD, let perpendiculars to it be drawn meeting a diameter AB in E and F; E and Fare equally distant from the centre 0. G E но B Draw OG perpendicular to CD, and therefore bisecting it; then OG is parallel to DF; whence GD: OF :: HG: HO :: HC: HE since the triangles HGO, HEC are equiangular; '.. (Eucl. v. 18, 15.) DG OF: GC: OE but GD=GC, .. OF= OE. (18.) If from the extremities of the diameter of a semicircle perpendiculars be let fall on any line cutting the semicircle; the parts intercepted between those perpendiculars and the circumference are equal. From A and B, the extremities of the diameter AB, let AC, BD be drawn perpendicular to any line CD cutting the semicircle in E and F; CE is equal to FD. A D G F From the centre draw OG perpendicular to CD, it will be parallel to AC and BD, whence CG: GD :: AO : OB, i. e. in a ratio of equality. But (Eucl. iii. 3.) EG = GF, and .. CE=FD. (19.) In a given circle to place a straight line parallel to a given straight line, and having a given ratio to it. Let AB be the given line in the cir A cle ABC whose centre is 0. Draw the diameter CD at right angles to AB: and E taking a line EF which has to AB the given ratio (Eucl. vi. 12.), place it in the circle ABC; bisect it in G and join OG; make OH=OG, and through H, draw IK parallel to AB; IK is the line required. For since OG=OH, .. (Eucl. iii. 14.) IK = EF, and EF : AB in the given ratio; .. IK : AB in the given ratio. (20.) Through a given point, either without or within a given circle, to draw a straight line, the part of which intercepted by the circle, shall be equal to a given line, not greater than the diameter of the circle. Let P be the given point without the circle ABC, whose centre is O. In the circle place a straight line AB equal to the given straight line; which bisect in E; and join OE. With the P B E centre O and radius OE describe a circle; this will touch AB in E, since the angles at E are right angles (Eucl. iii. 3.); from P draw PCD touching the circle in F. PCD is the line required. Join OF Then OF being equal to OE, CD will be equal to AB (Eucl. iii. 14.), i. e. to the given line. (21.) From a given point in the diameter of a semicircle produced, to draw a line cutting the semicircle, so that lines drawn from the points of intersection to the extremities of the diameter, cutting each other, may have a given ratio. Let C be the given point in the diameter BA produced. Make BC CD in the given ratio; and from the points E and E D, in which CD cuts the semicircle, draw EB, AD to the extremities of the diameter. CD is the line required. Since the angles EDA, EBA in the same segment are equal, and the angle at C common to the two triangles ACD, CEB, the triangles are equiangular, whence BE: AD: BC CD, i. e. in the given ratio. (22.) From the circumference of a given circle to draw to a straight line given in position, a line which shall be equal and parallel to a given straight line. Let AB be the given circle whose centre is O, and DE the line given in position. From O draw OF parallel and equal to the given line; and B D with the centre F, and radius equal to OB, the radius of the given circle, describe a circle cutting DE in G: join FG, and draw OA parallel to it; join AG; AG is the line required. = Since FG OB=OA, and is parallel to it, AG is equal and parallel to OF, and .. equal and parallel to the given line. (23.) The bases of two given circular segments being in the same straight line; to determine a point in it such, that a line being drawn through it making a given angle, the part intercepted between the circumferences of the circles may be equal to a given line. Let AB, CD, the bases of the segments be in the same line. Through O the centre of the circle ABI, draw EOG B making with AB an angle equal to the given angle, and make OE equal to the given line. From E draw EF, to the circle CFD, equal to the radius OB; draw OI parallel to EF; join IF cutting AD in H; H is the point required. For OI being equal and parallel to EF, OE is equal and parallel to IF, .. IF is equal to the given line; and IF being parallel to EG, the angle FHC is equal to EGB, i.e. to the given angle. If the distance of E from the centre of the circle CFD be less than the sum of the radii, there are two points in the circumference CFD, and two corresponding points in AD, which will answer the conditions. (24.) If two chords of a given circle intersect each other, the angle of their inclination is equal to half the angle at the centre which stands on an arc equal to the sum or difference of the arcs intercepted between them, according as they meet within or without the circle. Let AB, CD cut one other in the point E; and first within the circle ABC; the angle of inclination is equal to half the angle at the centre standing on an arc equal to the sum of CA and DB. Α F D B Through D draw DF parallel to BA. Find O the centre of the circle; join CO, FO. Then AB being parallel to FD, (ii. 1.) AF is equal to BD; and the angle CEA is equal to CDF, i. e. to half the angle COF, which stands on the arc CF equal to CA and BD together. 2. Next, let AB, CD intersect in E, without the circle. The same construction being made, the angle CEA is equal to the angle CDF, i.e. to half COF, i.e. to half the angle standing on E CF which is the difference between CA and AF, or CA and BD. (25.) If from a point without two circles which do not meet each other, two lines be drawn to their centres, which have the same ratio that their radii have; the angle contained by tangents drawn from that point towards the same parts will be equal to the angle contained by lines drawn to the centres. From the point A let the lines AB, AE be drawn to the centres of two circles, and let them have the same ratio that the radii BC, DE, have; from A draw the tangents AC, AD; as also AF, AG; each C B F E of the angles CAD, FAG will be equal to BAE. Since AB BC :: AE : ED, and the angles at C and D are right angles, .. the triangles ABC, ADE are |