Again (constr. and S. 26. 1.) the are equiangular, as are, also, the

gbh, GBH ich, ICH;

.. (E. 4. 6.) hg: hb:: HG: HB: And (constr.) hb: hc:: HB: HC: Also (E. 4. 6.) hc : hi :: HC: HI;

.. (E. 22, 5.) hg: hi:: HG: HI:

But (constr.) hg=hi; =hi; .·. HG=HI; and it is manifest, also, from the construction, that the GHIghi, of the square fghi; .. the 4 GHI is a right ▲.

Again, since (constr.) bh: hc:: BH: HC,

.. (comp. and div.) th: bh :: TH:BH: Lastly, (constr. and S. 26. 1.) the two tfh, TFH, are equiangular, as are, also, the two

BHG;

(E. 4. 6.) fh: th:: FH: TH:

And th: bh::TH:BH;

Also (E. 4. 6.) bh: hg:: BH: HG;
(E. 22. 5.) fh: hg:: FH: HG:

bhg,

Wherefore, the two fhg, FHG, having their sides about the equal a fhg, FHG, proportionals, are (E. 4. 6.) equiangular; .. the FGH is a

right angle; and (E. 4. 6.) FG = GH, because (constr.) fg = gh: And, as hath been shewn, the FGH, GHI, are right; .. (E. 28. 1.) GF is parallel to HI.

It has been shewn, also, that HI= HG; .. (E. 33. 1. and E. 34. 1.) the figure FGHI is equilateral and rectangular: That is (E. 30. def. 1.) it is a square.

41. PROBLEM.

PROP. XXXIII.

To determine the locus of the summits of all the triangles which can be described on a given base, so that each of them shall have its two sides in a given ratio.

Let AB be a given finite straight line: It is re

quired to determine the locus of the summits of all the which can be described upon AB, as a base, having their two remaining sides, in each, in a given ratio to one another.

Divide (E. 10. 6.) AB in C, so that AC shall be to CB in the given ratio; from the greater segment AC, cut off CD = CB; find (E. 11. 6.) a third proportional to AD and CB, and in AB, produced, make BK equal to it; from the centre K, at the distance KC, describe the circle CPE: The circumference CPE is the locus which was to be determined.

For, take any point P, in the circumference CPE, and draw PA, PB, PC, and PK: Then since,

(constr.) AD: CB:: CB: BK,

.. (E. 18. 5.) AD + CB or AC: CB:: CK: BK; .. (E. 16. 5.) AC: CK:: CB: BK;

(E. 18. 5.) AK: CK:: CK: BK;

i. e. (E. 15. def. 1.) AK:KP::KP:KB; ...(E. 6.6.) the two▲ APK, BPK, are equiangular: .. (E. 4. 6.) PA: PB:: AK: PK or CK: And it has been shewn that

AK: CK:: CK: BK:: AC: CB;

.. (E. 11. 5.) PA: PB:: AC: CB:

And (constr.) AC is to CB in the given ratio; .. PA is to PB in the given ratio, wherever, in the circumference CPE, the point P is taken.*

PROP. XXXIV.

42. PROBLEM. The base, the perpendicular distance of the vertex from the base, and the ratio of the two sides of a triangle being given, to construct it.

Draw (E. 31. 1. and E. 11. 1.) a straight line parallel to the given base, and at a perpendicular distance from it equal to the given perpendicular distance; draw, (S. 33. 6.) the

* If the given ratio be a ratio of equality, the locus to be determined is, manifestly, the straight line drawn at right angles to AB, through the point which divides AB into two equal parts.

locus of the summits of all the

which can be described on the given base, having their sides to one another in the given ratio; and it is manifest that the point, in which this locus meets the line drawn parallel to the base, will be the summit of the A which was to be de scribed.

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 (S. 83. 6.) the locus of the summits of all the A, 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. 11. 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 A 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 (S. 33. 6.) the locus of the summits of all 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 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.

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 (S. 33. 6.) the locus of the summits of all 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 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. 3. 6., that the point, in which the one locus cuts the other, is the point which was to be found.