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PROP. XXXV.

43. PROBLEM. The segments into which the perpendicular, drawn from the vertex to the base of a triangle, divides the base, and the ratio of the two remaining sides being given, to construct the triangle.

The segments being placed in the same straight line, upon their aggregate draw (Supp. xxxiii. 6.) the locus of the summits of all the triangles, which can be described on that line, as a base, so as to have their remaining sides in the given ratio: And it is evident that a perpendicular drawn (E. xi. 1.) to this base, from the point, which is common to the two segments, will cut the locus in a point, which is the vertex of the triangle that was to be described.

PROP. XXXVI.

44. PROBLEM. To find a point, from which if three straight lines be drawn to three given points, they shall be each to each in given ratios.

Upon the straight line joining two of the given points, describe (Supp. xxxiii. 6.) the locus of the summits of all the triangles having that line for a base, and having their sides to one another in one of the given ratios; upon the straight line, also, joining the third given point, and either of the other two, describe the locus of the summits of all triangles having that line for a base, and having their sides in another of the given ratios: Then it is manifest,

that the point, in which the one locus cuts the other, is the point which was to be found.

PROP. XXXVII.

45. PROBLEM. A straight line being divided into three given parts, to find a point without it, at which the three parts shall subtend equal angles.

Upon the aggregate of the first and second of the given parts, describe (Supp. xxxiii. 6.) the locus of the summits of all triangles, having that line for a base, and having their sides to one another, as the first is to the second of the given parts: Again, upon the aggregate of the second and third of the given parts, describe the locus of the summits of all triangles having that line for a base, and having their sides to one another as the second of the given parts is to the third: Then it is manifest, from E. iii. 6., that the point, in which the one locus cuts the other, is the point which was to be found.

PROP. XXXVIII.

46. PROBLEM. To find a point in a given line, from which, if two straight lines be drawn to two given points, both on the same side of the given line, they shall be to each other in a given ratio.

Upon the straight line joining the two given points, describe (Supp. xxxiii. 6.) the locus of the summits of all triangles having that line for a base, and having their sides in the given ratio; and it is evident, that the point, in which the locus, so

described, cuts the given line, is the point which was to be found.

PROP. XXXIX.

47. PROBLEM. In a given parallelogram to inscribe a parallelogram that shall have its two adjacent sides in a given ratio to one another, and that shall be the half of the given parallelogram.

Let ABCD be the given parallelogram: It is required to inscribe it in a parallelogram, which shall be the half of ABCD, and which shall have two adjacent sides in a given ratio to one another.

Bisect (E. x. 1.) AB in E, and through E draw (E. xxxi. 1.) EF parallel to AD or BC: And, first,

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if the given ratio be a ratio of equality, bisect, also, EF in K; through K draw (E. xi. 1.) LKM perpendicular to EF; and draw EL, LF, FM, and ME: Then ELFM is an equilateral parallelogram, and it is the half of the ABCD.

For (E. x. 6.) LM is divided, in K, in the same manner as AB is divided in E; .. KL=KM; therefore (constr. and E. iv. 1.) EL, and LF, and FM, and ME, are equal to one another; and therefore (Supp. xviii. 1.) the figure LEMF is a paral

lelogram And since (E. iv. 1.) the A ELF is the half of the AEFD, and the AEMF is the half of the EBCF, therefore the whole figure ELFM is the half of the given ABCD.

But, secondly, let the given ratio be not a ratio of equality In this case, upon EF describe (Supp. xxxiii. 6.) the locus of all the triangles having EF for a base, and having their sides in the given ratio, and let it cut AD in G; join E, G, and F, G; from CB cut off CH=AG, and join E, H and F, H: Then is EGFH the parallelogram which was to be described.

For (constr. and Supp. xliii. 1.) EGFH is a parallelogram; and it may be shewn to be the half of the given ABCD, in the same manner as ELFM was shewn to be half of ABCD; and (constr.) the adjacent sides EG and GF, are to one another in the given ratio.

PROP. XL.

48. PROBLEM. From a given point, either within or without a given rectilineal angle, to draw a straight line cutting the two lines which contain the angle, so that the distances of the two intersections from the given point, shall be to one another in a given ratio.

Let PAQ be the given rectilineal angle, and, first, let B be a point without it: It is required to draw from B a straight line cutting AP and AQ, so that the distances of its intersections with AP and AQ, from B, shall be to one another in a given ratio.

Through B draw BC to any point C-in AP;

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find (E. xii. 6.) a fourth proportional to the two straight lines, which exhibit the given ratio, and to BC; and from BC cut off BG equal to that fourth proportional; through G draw (E. xxxi. 1.) GE parallel to AC, and meeting AQ in E; join B, E and produce it to F: Then shall FB be to EB in the given ratio.

For (constr. and E. xxix. 1.) the two As BFC, BEG, are equiangular:

therefore (E. iv. 6.) FB: EB :: CB: GB:

But (constr.) CB is to GB in the given ratio; .. FB is to EB in the given ratio.

And, by the same method of construction, the problem may be solved, when the given point is within the given angle.

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49. COR. It is manifest that the problem admits of the same method of solution if one of the given lines, as AP, be a straight line of indefinite length, and if the other AQ be a line of any kind, in the same plane with AP.

PROP. XLI.

50. PROBLEM. To find, between two given parallel straight lines, the locus of all the points, from each of

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