1

is equal to DF (26. 1): But the square of AD, opposite the acute angle ACD of the triangle CAD, and twice the rectangle DCE are equal to the squares of AC CD (15. 2); and the square of CB, opposite the obtuse angle CDB of the triangle CBD, is equal to the squares of CD, DB, and twice the rectangle CDF [12. 2]; or, CE, DF being equal, as also AC, BD [34. 1], to the squares of AC, CD and twice the rectangle DCE; whence, adding equals to equals, the squares of AD, CB with twice the rectangle DCE, are equal to double the squares of AC, CD with twice the rectangle DCE; taking from each, twice the rectangle DCE, there remain the squares of AD and CB, equal to double the squares of AC and CD, or, AB, BD being severally equal to CD, CA, to the squares of AB, BD, CD and CA.

Cor. 2.-The squares of the sides [AC, AB] of a triangle [ACB], are double the squares of half the base [CB], and of a right line [AG], drawn to the middle [G] of the base, from the vertical angle [CAB].

Complete the parallelogram ACDB [Cor. 6. 34. 1]; the squares of its sides, being equal to the squares of its diagonals AD, CB [by the preced. cor.], the squares of AC and AB are equal to half the squares of AD, CB; but the squares of AD, CB are fourfold the squares of their halves [Cor. 4. 2], and AG, CG are the halves of AD, CB [Schol. 3. above]; therefore the squares of AC and AB, are double the squares of CG and AG.

PROP. XIV. PROB.

To constitute a square, equal to a given rectilineal figure (A).

Make the right angled parallelogram BCDE equal to A [45. 1]; and if its adjacent sides BE, ED are equal, it is a square, and what was proposed is done.

If not, on either of these sides as BE, produced, take EF equal to the other ED; bisect BF in G, and from the centro G, at the distance GB or GF, describe the semicircle BHF; let DE be produced to meet the semicircle in H; the square described on EH, is equal to the given rectilineal figure A.

For, drawing GH, because BF is bisected in G, and divided unequally in E, the rectangle BEF with the square of GE, is equal to the square of GF (5, 2); or, GH, GF being equal (Def. 10. 1), to the square of GH; and therefore (47. 1), to the squares of GE and EH; taking from each the common square of GE, the rectangle BEF is equal to the square of EH (Ax. 3. 1); but the rectangle BEF is equal to the rectangle BD, because ED is equal to EF, and so the square described on EH is equal to the rectangle BD, and therefore to the rectilineal figure A.

BOOK III.

DEFINITIONS.

1. A RIGHT line, is said to touch a circle, or to be a tangent to it, which meeting it, and being produced, does not cut it.

2. Circles, are said to touch one another, which meet, but do not cut each other.

3. A right line, is said to be inscribed in a circle, when its extremes are in the circumference of the circle.

4. Right lines, are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. See def. 21. Book 1.

5. And the right line, on which the greater perpendicular falls, is said to be more remote from the centre.

6. A segment of a circle, is a figure contained by a right line, and the part of the circum

ference it cuts off.

DE

7. The angle of a segment, is that which is included by the right line and the circumference.

8. An angle in a segment, is an angle, contained by two right lines, drawn from any point, in the part of the circumference, by which the segment is bounded, to its extremes.

9. An angle, is said to insist, or stand on, the part of the circumference (or arch), included between the legs of the angle.

10. A sector of a circle is, a figure, contained by two radiuses, and the part of the circumference between them.

11. Similar segments of circles, are such as receive equal angles.

PROPOSITION I. PROBLEM.

To find the centre of a given circle (ABC). Draw within the given circle any right line AB, which bisect in D (10. 1); from D, draw DC at right angles to AB (11. 1), which produce to meet the circumference in E; bisect EC in F. The point F is the centre of the circle.

F

C

D

No other point in EC, but F, can be the centre, for, if it were, the radiuses drawn from thence to C and E would be unequal, which is absurd (Def. 10. 1): if therefore F be not the centre, let some point, as G, without EC, be, if possible, the centre, and draw GA, GD, GB.

E

Because, in the triangles GDA, GDB, the side DA is equal to DB (Constr.), DG common, and GA equal to GB (Hyp. and Def. 10. 1), the angles GDA, GDB are equal (8. 1), and therefore right angles (Def. 20. 1); but the angle CDB is a right angle (Constr.), therefore the angles GDB, CDB are equal (Theor. at 11. 1), part and whole, which is absurd (Ax. 9. 1); therefore G is not the centre of the circle. In like manner it may be shewn, that no other point without EC is the centre of the circle; and it is above shewn, that no other point in EC, but F, is the centre; therefore F is the centre.

PROP. II. THEOR.

A right line, which joins any two points (A, B) in the circumference of a circle (ACB), falls wholly within the circle. If not, let AEB be a right line, of which a point E falls without the circle; find the centre of the circle D (1. 3), draw DE meeting the circumference in F, and join DA, DB.

Because, in the triangle DAB, the sides DA, DB are equal (Def. 10. 1), the angle DAB is equal to the angle DBA (5. 1); and the external angle DEB, of the triangle AED, is greater than the internal remote

angle DAE (16. 1), and therefore than its equal DBE, and so the side DB is greater than DE (19. 1); but DF is equal to DB [Def. 10. 1], therefore DF is greater than DE, the part than the whole, which is absurd: therefore the right line drawn from A to B does not in any part fall without the circle. In like manner it may be proved, that no part of it falls on the circumference, it falls therefore wholly within the circle.

PROP. III. THEOR.

If a right (CD), passing through the centre of a circle (ABC), bisect a right line inscribed in it (AB), not passing through its centre, it cuts it at right angles; and if it cut it at right angles, it bisects it.

Find the centre of the circle E [1. 3], and join EA, EB, which, being radiuses of the circle, are equal [Def. 10. 1], and the triangle EAB is isosceles [Def. 29. 1]; therefore if EF or CD bisect the base AB, it cuts it at right angles, and, if it cut it at right angles, it bisects it [Cor. 26. 1].

C

PROP. IV. THEOR.

Two right lines inscribed in a circle, cutting each other, and not passing both through the centre, do not bisect each other.

If one of the right lines pass through the centre, it is manifest, it is not bisected by the other, not passing through the centre.

But if neither of them, as AC, BD, pass through the centre, they cannot bisect each other; for let them, if possible, do it, and find the centre of the circle F (1. 3), and join EF; and since AC is bisected in E [Hyp.], FE is perpendicular to AC [3. 3], and the angle FEC right; and since BD is bisected in E, FE is perpendicular to BD [3. 3], and the angle FED right,