squares of CD, AD are equal to the square of AC (47. 1), and the squares of BD, AD to the square of AB (by the same); therefore the squares of BC, AC are equal to twice the rectangle BCD with the square of AB, and so the square of AB, is less than the squares of BC, AC, by twice the rectangle BCD.

Schol. 1.-The demonstration of the case of the 3rd or right hand figure, when the angle ABC is right, may also be thus : The square of AC is equal to the squares of AB, BC (47. 1); adding to each the square of BC, the squares of AC, BC are equal to the square of AB with twice the square of BC; and so the square of AB, is less than the squares of AC, BC, by twice the square of BC.

Schol. 2.-This proposition, the preceding, and the 47th of the 1st book, may be all comprised in one proposition, thus:The difference of the square of any side of a triangle, from the squares of the other two sides taken together, is equal to double the rectangle, under either of the sides including the angle opposite the` first mentioned side, and the segment of that side, produced, if necessary, between the same angle, and a perpendicular let falt thereon, from the opposite angle; the square of the first mentioned side being equal to, or greater or less than, the squares of the other two sides, according as the angle opposite thereto, is equal to, or greater, or less than, a right angle.

Schol. 3.-The diagonals (AD, CB) of a parallelogram bisect each other.

A

Let G be the point in which they meet; because the triangles AGB, DGC have the angles at G equal (15. 1), the angle BAGE

equal to its alternate CDG (29. 1),

B

and the sides AB, CD equal (34. 1); AG is equal to GD (26. 1). In like manner; CG, GB may be proved equal.

Cor. 1.-The squares of the diagonals (AD, CB) of a parallelogram (AD) taken together, are equal to the squares of all its sides (AB, BD, CD and CA).

On CD, produced as necessary, let fall the perpendiculars AE, BF (12. 1); and in the triangles ACE, BDF, the angles CEA, DFB are equal, being right angles, and, because of the parallels AC, BD, the external angle BDF is equal to the internal remote ACE (29. 1), and the sides AC, BD opposite the right angles at E and F, are equal (34. 1); therefore CE

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 (13. 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 centre 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 circumference it cuts off.

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.

T

C

D

G

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 ᎠᎪ, ᎠᏴ.

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

E

F