COMPOSITION. Let fall the perpendicular AD, which divide at E in the given ratio, and erect the perpendicular CE; this straight line is the locus required. For CE being parallel to BD, AC: AB:: AE: AD, that is, in the given ratio. PROP. II. THEOR. If a straight line, drawn through a given point to the circumference of a given circle, be divided in a given ratio, the locus of the point of section will also be the circumference of a given circle. Let AB, terminating in a given circumference, be cut in a given ratio; the segment AC will likewise terminate in a given circumference. ANALYSIS. B Join A with D the centre of the given circle, and draw CE parallel to BD. It is obvious (VI. 1. El.) that AC: AB :: AE: AD; whence the ratio of AE to AD being given, AE and the point E are given. Again, since (VI. 2. El.) AC: AB :: CE: BD, the ratio of CE to BD is given, and consequently CE is given in magnitude. Wherefore the one extremity E being given, the other extremi A ED ty of CE must trace the circumference of a given circle, COMPOSITION. Join AD, and divide it at E in the given ratio, and in the same ratio make DB to the radius EC, with which, and from the centre E, describe a circle. For draw AB cutting both circumferences, and join CE and BD. Because CE: BD:: AE: AD, alternately CE: AE :: BD: AD; wherefore the triangles CAE and BAD, having likewise a common angle, are similar, and consequently AC: CB:: AE: AD, that is in the given ratio. PROP. III. THEOR. If, through a given point, two straight lines be drawn in a given ratio and containing a given angle; if the one terminate in a straight line given in position, the other will also terminate in a straight line given in position. Let the ratio of BA to AC, with the angle BAC and its vertex A, be given; if the extremity B lie in the straight line BD, the extremity C will have for its locus another straight line given likewise in position. ANALYSIS. Let fall the perpendicular AD upon BD, draw AE forming with AD an angle DAE equal to BAC, and make AB: AC :: AD: AE; CE being joined, is the locus required. Because the angle DAE is, by construction, equal to BAC, it is given; and the perpendicular AD being given, the straight line AE is, therefore, given in position. But AB: AC:: AD: AE, and this being a given ratio, AE is hence given also in magnitude. Again, since the angle BAC is equal to DAE, the angle BAD is equal to CAE; and because AB: AC AD: AE, alternately AB: AD :: AC : AE; wherefore the triangles ABD and ACE, having their vertical angles equal, and the sides containing those angles proportional, are (VI. 15. El.) similar, and consequently the angle CEA is equal B E C to BDA, and therefore a right angle; consequently the straight line EC is given in position. COMPOSITION. Having let fall the perpendicular AD, and made the angle DAE equal to BAC, make AD to AE in the given ratio, and, at right angles to AE, draw EC; this is the locus required. For the triangles BAD and CAE, having their vertical angles equal, and the angles at D and E right angles, are similar, and consequently AB: AD :: AC: AE, or alternately AB AC AD AE, that is, in the given ratio. : PROP. IV. THEOR. If, through a given point, two straight lines be drawn in a given ratio, and containing a given angle; if the one terminate in a given circumference, the other will also terminate in a given circumfe rence. Let the angle BAC, its vertex A, and the ratio of its 10 sides, be given; if AB be limited by a given circle, the locus of C will also be a given circle. ANALYSIS. Join A with D the centre of the given circle, draw AE at the given angle with AD, and in the given ratio, and join DB and EC. Because the point A and the centre D are given, the straight line AD is given; and since the angle DAE, being equal to BAC, is given, AE is given in position. But AD being to AE in the given ratio, AE must be given also in magnitude, and conse quently the point E is given. Again, the whole angle BAC being equal to DAE, the part BAD is equal to CAE; and because AB: AC :: AD: AE, alternately AB: AD: AC: AE; wherefore the triangles ADB and AEC are similar, and hence AB: BD:: AC: AE, or alternately AB: AC :: BD: CE; E D consequently the fourth term CE is given in magnitude; and its extremity E being given, the other must lie in a given circumference. COMPOSITION. Having drawn AE at the given angle with AD, make AD to AE in the given ratio, and in the same ratio let DB be made to EC; a circle described from the centre E with the distance EC, is the locus required. For AD AE: DB: EC, and alternately AD: DB:: AE: EC; but the angle BAD is equal to CAE, because the whole BAC is equal to DAE; consequently the tri. angles ABD and ACE are similar, and AB; AD :: AC: AE or alternately AB: AC:: AD: AE, that is, in the given ratio. PROP. V. THEOR. The middle point of a given straight line which is placed between the sides of a right angle, lies in the circumference of a given circle. Let AB be placed in the right angle EDF, the locus of its bisection C is a given circle. ANALYSIS. Join DC. Then (II. 30. El.) because the base of the From D, with a distance equal to half the given line, describe a circle; this is the locus required. For draw the radius DC, make AC=DC and produce 1 |