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which if two straight lines be drawn to the two given parallels, so as always to make with them, towards the same parts, given angles, they shall be to one another in a given ratio.

Let VW and XY be the two given parallels; let

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AB and AC, drawn from any point A in XY, be in the two given directions: It is required to find, between VW and XY, a locus, from any points of which if two straight lines be drawn to VW and XY, the one parallel to AC and the other parallel to AB, they shall be to one another in a given ratio.

Find (E. xii. 6.) a fourth proportional to the two straight lines, which exhibit the given ratio, and to AB; and from CA, produced, cut off AD equal to that fourth proportional; join B, D; through A draw (E. xxxi. 1.) AE parallel to DB, and through E draw EF parallel to CA, and let it meet AB in F; lastly, through F draw SFT parallel to VW or to XY: Then is ST the locus which was to be found. For take any point P, in ST, and from P draw PQ parallel to AB, and PR parallel to AC.

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And since (constr. and E. xxix. 1.) the as AEF, ABD are equiangular,

therefore (E. iv. 6.) FA : FE :: AB : AD: But (constr. and E. xxxiv. 1.) FA= PQ, and FE= PR; also (constr.) AB is to AD in the given ratio; .. PQ is to PR in the given ratio.

PROP. XLII.

51. PROBLEM. To divide a given straight line into two parts, such, that the rectangle contained by the whole line and one of its parts, shall have a given ratio to the square of the other part*.

Let AB be the given straight line: It is required

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to divide it into two parts, such that the rectangle contained by AB and one of the parts shall have to the square of the other part a given ratio.

Find (E. xii. 6.) a fourth proportional, L, to the two straight lines, which exhibit the given ratio, and to AB; and divide (Supp. lxxxi. 3.) AB into two parts, in C, so that AC × L = CB2: And since, (E. i. 6.) AC×AB: AC×L or CB2 :: AB : L, it is manifest that AB has been divided in C, so that AC × AB is to CB' in the given ratio.

This is the general problem, of which E. xxx. 6. is a particular case.

PROP. XLIII.

52. PROBLEM. One given circle lying within another, to find a point from which, if two tangents be drawn, one to each of the given circles, they shall be to each other in a given ratio.

Let ABC, DEF, be two circles, of which

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DEF lies within ABC: It is required to find a point from which if tangents be drawn to touch the two circles ABC, DEF, they shall be to one another in a given ratio.

Draw (E. xvii. 3.) AL touching the lesser circle DEF in any point E, and let AL meet the circumference of ABC in A and B; bisect (E. x. 1.) AB in G, and from E draw (E. xi. 1.) EH perpendicular to AB; find (E. xii. 6.) a fourth proportional to the two straight lines, which exhibit the given ratio, and to AG; and make EH equal to it; from the centre G, at the distance GA or GB, describe the circle AKB; from H draw (E. xvii. 3.) HK touching the circle AKB in K; and produce HK to meet AB, produced, in L: Then is L the point which was to be found.

For, from L draw LC touching the circle ABC in C; and join G, K; therefore (constr. and E. xviii. 3.) the GKL is a right angle, as is, also, (constr.) the ▲ LEH; therefore (Supp. xvi. 1.) the as LKG, LEH, are equiangular;

therefore (E. iv. 6.) LE : LK :: EH : GK or GA: But, since (E. xxxvi. 3.) AL × LB is equal to LK2, and also to LC2; .. LK=LC; and (constr.) EH is to GA in the given ratio; therefore the tangent LE is to the tangent LC in the given ratio.

PROP. XLIV.

53. PROBLEM. From a given point, to draw a straight line to cut a given circle, so that the distances of the two intersections from the given point, shall be to each other in a given ratio.

Let CFE be the given circle, and the given

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point without it: It is required to draw from A a straight line cutting CFE, so that the distances of its two intersections from A shall be to one another in a given ratio.

From A draw (E. xvii. 3.) AF touching the circle CFE in F; find (Supp. xxi. 6.) a square, which shall

be to the square of AF, in the given ratio; from the centre A, at a distance equal to the side of the square thus found, describe a circle cutting the circumference of CFE in Q; and draw AQ, which is, therefore, equal to the side of that square; produce AQ to meet the circumference of CFE again in P: Then shall AP be to AQ in the given ratio.

For (E. i. 6.) AP : AQ :: AP × AQ : AQ2: But (E. xxxvi. 3.) AP × AQ = AF2; and (constr.) AF2 is to AQ2 in the given ratio; .. AP is to AQ in the given ratio.

PROP. XLV.

54. PROBLEM. Two given circles lying wholly without one another, through a given point, which is between the two circles, and which is posited in the straight line joining their centres, to draw a straight line that shall be terminated by the convex circumferences, and divided, by the given point, into two parts, that are to one another in a given ratio.

Let BD and EF be two circles, and A a

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KH

F

point in CK, which joins the two centres C and K: It is required to draw, through A, a ́straight line, which being terminated by the convex circum

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