A

M

angles BCF, DCF, the two sides BC, CF are equal to the two, DC, CF; and the angle BCF is equal to the angle DCF; therefore the base BF is equal (4. 1.) to the base FD, and the other angles to the other angles, to which the equal sides are opposite; therefore the angle CBF is equal to the angle CDF:

B

H

c

K

D

And because the angle CDE is double of CDF and that CDE is equal to CBA, and CDF to CBF: CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bisected by the straight line BF:

In the same manner it may be demonstrated, that the angles BAE, AED are bisected by the straight lines AF, FE.

From the point F, draw (12. 1.) FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA :

And because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC, in the triangles FIIC, FKC there are two angles of one equal to two angles of the other; and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore the other sides shall be equal (26. 1.), each to each; wherefore the perpendicular FH is equal to the perpendicular FK:

In the same manner it may be demonstrated that FL, FM, FG are each of them equal to FH, or FK;

Therefore the five straight lines FG, FH, FK, FL, FM are equal to one another; wherefore the circle described from the centre F, at the distance of one of these five, shall pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right angles, and that a straight line drawn from the extremity of the diameter of a circle at right angles to it touches (16. 11.) the circle; therefore each of the straight lines AB, BC, CD, DE, EA touches the circle: wherefore it is inscribed in the pentagon ABCDE. Which was to be done.

PROPOSITION XIV.

PROB. To describe a circle about a given equilateral and equiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to describe a circle about it.

A

Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E.

It may be demonstrated, in the same manner as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA, FE:

And because the angle BCD is equal to the angle CDE, and that FCD is the half of the

B

C

angle BCD, and CDF the half of CDE; the angle FCD is equal to FDC; wherefore the side CF is equal (6. 1.) to the side FD:

In like manner it may be demonstrated that FB, FA, FE are each of them equal to FC or FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another; and the circle described from the centre F, at the distance of one of them, shall pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done.

PROPOSITION XV.

PROB. To inscribe an equilateral and equiangular hexagon in a given circle.

Let ABCDEF be the given circle; it is required to inscribe an equilateral and equiangular hexagon in it.

Find the centre G of the circle ABCDEF, and draw the diameter AGD; and from D as a centre, at the distance DG, describe the circle EGCH; join EG, CG, and produce them to the points B, F; and join AB, BC, CD, DE, EF, FA: the hexagon ABCDEF is equilateral and equiangular.

Because G is the centre of the circle ABCDEF, GE is equal to GD:

And because D is the centre of the circle EGCH, DE is equal to DG;

Wherefore GE is equal to ED, and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, be- F cause the angles at the base of an isosceles triangle are equal (Cor. 5. 1.); and the three angles of a triangle are equal (32. 1.) to two right angles; therefore E the angle EGD is the third part of two right angles : In the same manner it may be demonstrated, that the angle DGC is also the third part of two right angles:

H

B

And because the straight line GC makes with EB the adjacent angles EGC, CGB equal (13. 1.) to two right angles, the remaining angle CGB is the third part of two right angles; therefore the angles EGD, DGC, CGB are equal to one another; and to these are equal (15. 1.) the vertical opposite angles BGA, AGF, FGE; therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another:

But equal angles stand upon equal (26. 111) circumferences; therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one

another :

And equal circumferences are subtended by equal (29. 111.) straight lines; therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular; for, since the circumference AF is equal to ED, to each of these add the circumference ABCD; therefore the whole circumference FABCD shall be equal to the whole EDCBA: and the angle FED stands upon the circumference FABCD, and the angle AFE upon EDCBA ; therefore the angle AFE is equal (27. 111.) to FED: in the same manner it may be demonstrated, that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED; there

fore the hexagon is equiangular; and it is equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Which was to be done.

COR. From this it is manifest, that the side of the hexagon is equal to the straight line from the centre, that is, to the semidiameter of the circle.

And if through the points A, B, C, D, E, F, there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonstrated from what has been said of the pentagon: and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon.

PROPOSITION XVI.

PROB. To inscribe an equilateral and equiangular quindecagon in a given circle.

Let ABCD be the given circle; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD.

Let AC be the side of an equilateral triangle inscribed (2. Iv.) in the circle, and AB the side of an equilateral and equiangular pentagon inscribed (11. Iv.) in the same;

Therefore, of such equal parts as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third part of the whole, contains five;

B

C

And the circumference AB, which in the fifth part of the whole, contains three;

Therefore BC, their difference, contains two of the same parts:

Bisect (30. III.) BC in E: therefore BE, EC are each of them the fifteenth part of the whole circumference ABCD; therefore, if the straight lines BE, EC be drawn and straight lines equal to them be placed round (1. IV.) in the whole circle, an equilateral and equiangular quindecagon shall be inscribed in it. Which was to be done.

And in the same manner as was done in the pentagon, if, through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon shall be described about it: and likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it.

EXERCISES ON BOOK IV.

1. In a given circle to inscribe a line of given length, so as :

(a) To pass through a given point;

(b) To touch a given circle;

(c) To make a given angle with a given line;

(d) To cut off one-third of another given circumference.

In all cases the limits of possibility to be assigned.

2. If all possible triangles having the same base and equal vertical angles be supposed to exist, show that the centres of the inscribed and escribed circles are situated upon the circumferences of determinable circles.

3. Describe three circles, each touching the other two, and the three centres being three given points, in all possible ways.

4. Describe a circle which shall pass through a given point and touch a given line, and a given circle; and assign the limits of possibility.

5. From the three angles of a triangle draw perpendiculars to the opposite sides; and join the intersections two and two: then

(a) The perpendiculars intersect in the same point;
(b) They bisect the angles of the second triangle;

(c) The circle about the second triangle bisects the sides of the first;

(d) The diameter of this circle is half that described about the first triangle.

6. In the figure to IV. 10., produce DC to meet the circle in F, and show that:

:

(a) AC is the side of a regular pentagon inscribed in the smaller circle ACD;

(b) Of a regular decagon, in the larger one BED;

(c) The angle ABF will be triple of BFD.

7. If there be drawn lines to the alternate angles of a regular pentagon, their intersections will give the angular points of another regular pentagon; and if the alternate sides be produced, their intersections will be the angular points of a third regular pentagon; and the circles circumscribing and inscribed in these pentagons will all be concentric.

8. Divide a right angle into three, four, five, six, eight, ten, twelve and fifteen equal parts.

9. Every regular polygon may have a circle inscribed within it, and another circumscribed about it; and these circles will be concentric.

10. An equilateral figure inscribed in a circle, or circumscribed about it, is equiangular; and an equiangular figure so inscribed or circumscribed is equilateral.

11. A triangle being constituted of the sides of the regular pentagon, hexagon and decagon inscribed in the same circle, will be right angled; and the same is true of the regular inscribed triangle, square, and hexagon.

12. Given the radius of the inscribed circle and the perimeter of the triangle, to construct it in the cases where the third datum is :

(a) The sum of the sides;

(b) The difference of the sides;

(c) The perpendicular on the base;

(d) The radius of the circumscribing circle;
(e) The vertical angle;

(f) The difference of the angles at the base.

13. If there be equilateral triangles described externally on the three sides of any triangle, the three distances between the vertices of these and the opposite angles of the original triangle will be all equal:

And if the original triangle be right-angled, two of the equilateral triangles will together be equal to the third.

14. A regular polygon and a circle concentric with it being given, and from any point in the circle draw lines to all the angles, and likewise lines perpendicular to the sides: the sum of the perpendiculars is constant; and the sum of the squares on all the other lines is constant, wherever in the circumference the point be taken.