Let ABCD be any quadrilateral inscribed in a circle, and let AC, DB be drawn; then

Again, because the angle ABE is equal to the angle DBC, and the angle BAE to the angle BDC in the same segment, the triangle ABE is equiangular to the triangle DBC.

But we have shown BC. DA=BD. EC;

BA.DC+BC.DA=BD.AE+BD. EC.

But (II. 1) BD.AE+BD. EC-BD. AC;

BA.DC+BC.DA=BD.AC.

I

EXERCISES.

1. To draw a line from the vertex of a triangle bisecting the triangle (V. 1, Cor. 2).

2. Two squares are to each other as the squares of their diagonals.

3. If a line be drawn parallel to the base of a triangle, all lines from the vertex to the base will be cut proportionally; and the segments of the base and of the parallel will be proportional to each other.

4. The perimeters of similar polygons are proportional to their homologous sides (IV. 16).

5. A line drawn parallel to the parallel sides of a trapezoid will cut the other sides proportionally.

6. If two triangles be on equal bases and between the same parallels, any straight line parallel to their bases will cut off equal triangles and trapezoids.

7. Let A, B and C be three points in a straight line; and through C let any straight lines pass; the perpendiculars on these lines from A and B bear a constant ratio to each other.

8. If one of the parallel sides of a trapezoid be double the other, its diagonals trisect each other.

9. Trisect a given line by use of (V. 3).

10. In the figure in (II. 11) show that four other lines besides the given one are divided in extreme and mean

ratio.

is

11. The area of an equilateral triangle whose side is 1

V3

4

12. If the arc AB be bisected in C, and D be any point of the circumference, then AD+DB: DC:: AB: AC (V. 30).

13. A circle is described about an equilateral triangle; from any point of the circumference straight lines are drawn to the

D

angles of the triangle; then one of these lines is equal to the sum of the other two.

14. On a given line to construct a rectangle which shall be equal to a given rectangle (V. 14).

15. If from a point without the circle two lines be drawn, one cutting the circle and the other meeting, and the rectangle contained by the whole cutting-line and the part without the circle be equal to the square of the line meeting it, the latter line touches the circle (V. 27).

16. The parallelograms about the diagonal of a parallelogram are similar to the whole.

17. The base of a triangle is 6; of the triangle are cut off toward the base by a line parallel to it; what is the length of the cutting-line (V. 19)?

18. The area of a triangle is equal to its perimeter multiplied by the radius of the inscribed circle.

For other practical problems in this book see Mensuration at the end of Book VI.

BOOK VI.

REGULAR POLYGONS.-MEASUREMENT OF THE

CIRCLE.

DEFINITIONS.

1. A regular polygon is one which is equilateral and equiangular.

2. A polygon of five sides is called a pentagon; of six sides, a hexagon; of eight sides, an octagon; of ten sides, a decagon; of fifteen sides, a pentedecagon, etc., etc.

3. The apothem is the line drawn from the centre of a regular polygon at right angles to one of its sides.

Proposition 1.

Problem. To inscribe a square in a given circle.

Let ABCD be the given circle; it is required to inscribe a square in ABCD.

Draw the diameters AC, BD at right angles to each other, and join AB, BC, CD, DA; because in the triangles BEA, DEA, BE, EA and the angle BEA are equal to DE, EA and the angle DEA, each to each; therefore (I. 7) BA is equal to AD; and, for the same reason, BC, CD are each of them equal to BA or AD; therefore the quadrilateral figure

132

B

E

D

ABCD is equilateral. It is also rectangular, for the straight line BD being a diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right angle (III. 21); for the same reason each of the angles ABC, BCD, CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular; therefore it is a square, and it is inscribed in the circle ABCD.

Scholium. Since the triangle AED is right-angled and isosceles, we have (I. 42, Cor. 2)

AD: AE :: V2 : 1, or AD= AE√2.

Proposition 2.

Problem.-To describe a square about a given circle.

Let ABCD be the given circle; it is required to describe a square about it.

G

A

F

E

B

Draw two diameters, AC, BD, of the circle at right angles to each other, and through the points A, B, C, D draw (III. 15) FG, GH, HK, KF, touching the circle. Hence the angles at A, B, C, D are right angles (III. 14, Cor. 2); and because the angle A EB is a right angle, as likewise is EBG, GH is parallel (I. 26) to AC; for the same reason AC is parallel to FK, and, in like manner, GF, HK may be shown to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms, and GF is therefore equal (I. 32) to HK, and GH to FK; and because AC is equal to BD, and also to each of the two GH, FK, and BD to each of the two GF, HK, therefore GH, FK are each of them equal to GF or HK; therefore the quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB (I. 32)

H

C

K