B

E

EBC; and (E. 21. 3.) the

BDA= 2 BCE;

.. (E. 32. 1.) the ABD, BCE, are equiangular; .. (E. 4. 6.) BC:CE:: BD: DA;

.. (E. 16. 6.) BC X AD=BDX CE: Again, because (constr.) the ABE = ▲ DBC, and (E. 21. 3.) the 4 BAE BDC, the ABE,

BCD, are equiangular;

.. (E. 4. 6.) BA:AE::BD: DC;

.. (E. 16. 6.) BA × DC = BD × AE: And it has been shewn that BCX AD=BD × CE; (E. 1. 2.) AC× BD = AB × CD + AD ×

BC.

PROP. LVIII.

67. THEOREM. If, from the centre of the circle, described about a given triangle, perpendiculars be drawn to the three sides, their aggregate shall be equal to the radius of the circumscribed circle, together with the radius of the circle inscribed in the given triangle.

Let ABC be the given A; bisect (E. 10. 1.)

AB, BC, and AC in the points D, E, and F; and from D, E, and F draw (E. 11. 1.) DG 1 to AB, EG to BC, and FG to AC; then (S. 4. 1.) these perpendiculars meet in the same point G, which is the centre of the circle that can be described about the A ABC; find, also, (E. 4. 4.) the centre K, and the semi-diameter KH, of the circle that can be inscribed in the ▲ ABC; and draw GA: Then* GD + GE + GF GA + KH.

=

For draw DE, EF, and FD, .. (S. 69. 1. cor. 1. and E. 34. 1.) AC, CF = AB, and FD=X } BC; draw GB, and GC; And, since (constr.) the

at D, E, F, are right,.. (E. 32. 1. cor. 1.) the two DAF, DGF, are, together, equal to two right;.'. (S. 28. 3.) a circle may be described about the trapezium ADGF; and in the same manner it may be shewn that circles may be described about BDGE, and CFGE:

.. (S.57.6.) AGXDF+BGXDE+CGXFE=

* The straight lines GD, GE, GF, are wanting in the figure.

AFX DG+ADXGF+BDXGE+BEX DG+ CEXGF+CFXGE: And if to the doubles of these equals be added the rectangles GEX BC +GF× AC+GD × AB, which (E. 41. 1.) make up the double of the AABC, it will be manifest, from E. 1. 2., that the rectangle contained by the perimeter of the ▲ ABC, and by GA, together with the double of the ▲ ABC, is equal to the rectangle contained by the perimeter of ABC, and by the aggregate of GD, GE, and GF: But (S. 2. 4.) the double of the ▲ ABC is equal to the rectangle contained by the perimeter of the A and the semi-diameter, KH, of the circle inscribed in it; .. (E. 1. 2.) the rectangle contained by the perimeter, and by the aggregate of GA and KH, is equal to the rectangle contained by the perimeter, and by the aggregate of GD, GE, and GF; .. GD+GE+ GF GA + KH.

PROP. LIX.

68. PROBLEM. To find a point, from which if three straight lines be drawn to three given points, their differences shall be severally equal to three given straight lines; the difference of any two of the straight lines to be drawn, not being greater than the distance of the two points to which they are to be drawn.

Let A, B, C, be the three given points, and R, S,

two of the given differences: It is required to find a point, from which if three straight lines be drawn to A, B, and C, the difference of the first and second shall be equal to R, the difference between the second and third equal to S, and ... the difference between the first and third equal to the third of the given differences.

Draw AB, BC, and CA; bisect (E. 10. 1.) AB in D, and BC in E; from DB cut off DF, equal to a third proportional (E. 11. 6.) to 2AB, and to S; likewise from EB cut off EG, equal to a third proportional to 2BC, and to R; and through F and G draw (E. 11. 1.) FH 1 to AB, and GH to BC, and let them meet in H; find (E. 12. 6.) a fourth proportional, (T,) to AB, S, and BC; through H draw (S. 7. 6.) the locus, IH, of all the points, from which if perpendiculars be drawn to AB and BC, respectively, they shall cut off from GB and FB segments that are to one another as R is to T; lastly, in IH find (S. 96. 3.) a point K, such that the difference of its distances from C

and B, shall be equal to R: Then is K the point which was to be found.

For if not, let P be the point; and, if it be possible, let the point P be out of IH; join P, A, and P, B, and P, C; and draw, from P (E. 12. 1.) PL 1 to AB, and PM 1 to BC: Then (constr. E. 17. 3. and S. 96. 3. cor. 1.)

FLX AB = BPX5; and GMX BC=BPXR; .. (E. 16. 6. and constr.)

BP: FL:: AB:S:: BC:T;

and GM: BP::R: BC;

.. (E. 23. 5.) GM: FL:: R:T:

...(constr. and S. 7. 6. cor.) the point P cannot be out of IH; .. (constr. and S. 96. 3. cor. 2.) K is the point which was to be found.

PROP. LX.

69. PROBLEM. To describe a circle, which shall pass through a given point, and touch two given circles.

Find a point (S. 59. 6.) such that the difference between its distance from the centre of the one circle, and its distance from the given point, shall be equal to the semi-diameter of that circle; and that the difference between its distance from the other centre, and from the given point, shall likewise be equal to the other given semi-diameter: