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PROPOSITION 16.-PROBLEM.

B

EN

To inscribe a regular quindecagon in a given circle.

(References—Prop. III. 27, 30; IV. 1, 2, 11.) Given.—Let ABCD be the given circle.

Sought. --It is required to inscribe a regular quindecagon in it.

Construction.—1. Let AC be the side of an equilateral triangle inscribed in the circle (IV. 2), and AB the side of a regular pentagon inscribed in the same. (IV. 11.)

2. Bisect the arc BC in E (III. 30), and join BE, EC.

3. Draw straight lines equal to BE or EC, and place them round in the whole circle, contiguous to one another. (IV.1.)

Then shall a regular quindecagon be inscribed in the circle.

Proof.-1. Because of such equal parts as the whole circumference ABCDF contains fifteen; the arc ABC, which is the third part of the whole, contains five; and the arc AB, which is the fifth part of the whole, contains three;

2. Therefore their difference, the arc BC, contains two of

3. But BC was bisected in E; therefore BE, EC, are each of them the fifteenth part of the whole circumference ABCD.

4. Therefore the inscribed quindecagon is equilateral.

5. And because each of its angles stands upon the thirteenfifteenths of the circumference, it is also equiangular. (III. 27.)

Conclusion.—Therefore, a regular quindecagon has been inscribed in the circle ABCD. Q. E. F.

Corollary.-In the same manner, as was done for the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, a regular quindecagon will be described about it; and also, as for the pentagon, a circle may be inscribed in a given regular quindecagon, and circumscribed about it.

the parts.

EXERCISES ON BOOK IV.

Prop. 1 to 9. 1. In a given circle, to place a straight line equal to a given line, and making with another given line a given angle.

2. If a triangle be inscribed in one of two concentric circles, iņscribe another equiangular to it, in the other circle, and having its sides respectively perpendicular to those of the first triangle.

3. The square of the side of an equilateral triangle inscribed in a circle is equal to three-fourths of the square of the dia: meter.

4. The side of an equilateral triangle inscribed in a circle is equal to the perpendicular of an equilateral triangle described on the diameter.

5. If tangents be drawn through two of the angular points of an equilateral triangle inscribed in a circle, a diameter drawn from the third angular point will bisect the angle made by these tangents; and will be itself trisected by the circumference.

6. The diameter of the circle inscribed in a right-angled triangle is equal to the difference between the hypotenuse and the sum of the two legs.

7. Given the three sides of a triangle, equal respectively to 45, 39, 42, calculate the radius of the inscribed circle.

8. Given the vertical angle of a triangle, the line bisecting it, and the radius of the inscribed circle, to describe the triangle. 9. Inscribe a square in a given semicircle.

Prop. 10 to 16. 10. On a given straight line as base, describe an isosceles triangle, having each of the angles at the base one-third of the vertical angle.

11. The square of one of the diagonals of a regular pentagon inscribed in a circle, is less than the square of the diameter by the square of the side of the inscribed decagon.

12. ABCDE is a regular pentagon, and AC, BD, are joined by lines which intersect in F; prove that AFDE is a rhombus; and BFC an isosceles triangle having its vertical angle treble of either of the angles at the base.

13. A regular hexagon inscribed in a given circle is equal to three-fourths of the one described about the same circle.

14. A regular hexagon inscribed in a circle is double of an equilateral triangle inscribed in the same circle.

15. On a given straight line describe a regular pentagon, and also a regular hexagon.

16. Divide a right angle into five, and also into fifteen equal parts.

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