double of the squares of the bisecting line AC, and of half the side subtending the angle.

If the bisecting line be perpendicular to the side, it is manifest from Prop. 47. B. 1.

If not, draw from the angle A the line AF perpendicular to the opposite side; since one of the angles ACB and ACP is obtuse, let ACB be the obtuse. angle, and the square of AB is equal to the sum of the squares of AC and CB, together with twice the (1) Prop. 12. rectangle under BC and CF (1): the angle ACP is acute, and therefore the square of AP, together with twice the rectangle under PC and CF, or (as PC (2) Hypoth. and BC are equal (2) with twice the rectangle under BC and CF, is equal to the sum of the squares of (3) Prop.13. AC and CP (3); therefore the squares of AB and AP, with twice the rectangle under BC and CF, are equal to twice the square of AC, with the squares of BC and CP, and twice the rectangle under BC and CF: take away from both twice the rectangle under BC and CF, and the sum of the squares of AB and AP is equal to twice the square of AC, with the squares of BC and CP, or with twice the square of BC; because CP and BC are equal.

Fig. 7.

Schol. Hence it is evident that the sum of the squares of the sides of a parallelogram is equal to the (1) Prop.29. sum of the squares of the diagonals.

Because the angles DBC and BDA, ACB and CAD are equal (1), and the side BC is equal to AD (2), the diagonals mutually bisect each other (3), therefore the sum of the squares of the sides is equal to four times. the squares of the halves of the diagonals (4), or to the sum of the squares of the diagonals (5).

PROP. XIV. PROB.

Fig. 28.

To make a square equal to a given rectilineal figure

(Z).

Make a rectangle CI equal to the given rectilineal (1)Prop. 45. figure (1): if the adjacent sides be equal, the problem is done.

B. 1.

(2) Prop. 3. B. 1.

If not, produce either side IA and make the produced part AL equal to the adjacent side AC (2);

bisect IL in O, and from the centre O, with the radius OL, describe a semicircle LBI, and produce CA till it meet the periphery in B: the square of AB is equal to the given rectilineal figure.

For draw OB, and because IL is bisected in O and cut unequally in A, the rectangle under IA and AL, together with the square of OA, is equal to the square

of OL (3), or of OB which is equal to OL, and there- (3) Prop. 5. fore to the squares of OA and AB (4); take away from (4) Prop.47. both the square of OA, and the rectangle under IA B.1. and AL is equal to the square of AB: but the rectangle under IA and AL is equal to IC, for AL and AC are

equal (5); therefore the square of AB is equal to the (5) Constr. rectangle IC, and therefore to the given rectilineal figure Z.

E 2

THE

ELEMENTS OF EUCLID.

BOOK III.

See N.

Plate 3.
Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

See N.

DEFINITIONS.

1. EQUAL circles are those, whose diameters are equal.

2. A right line is said to touch a circle, when it meets the circle, and being produced does not cut it.

3. Circles are said to touch one another, which meet, but do not cut one another.

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.

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

6. A segment of a circle is the figure contained by a right line, and the part of the circumference it cuts off.

7. An angle in a segment is the angle contained by two right lines drawn, from any point in the circumference of the segment, to the extremities of the right line, which is the base of the segment.

8. An angle is said to stand upon the part of the circumference, or the arch, intercepted between the right lines that contain the angle.

9. A sector of a circle is the figure contained by two radii and the arch between them.

10. Similar segments of circles are those, which contain equal angles.

PROP. I. PROB.

To find the centre of a given circle (ACB).

Fig. 5.

B. 1.

(2) Prop.11.

Draw within the circle any right line AB, bisect it (1) Prop. 10. in D (1); from D draw DC perpendicular to AB (2), and produce it to E; bisect CE in F, and Fis the centre. For, if it be possible, let any other point G be the centre; and draw GA, GD, and GB.

B.1.

& Def. 15. B. 1.

(4) Constr. (5) Prop. 8.

Because in the triangles GDA, GDB, the side GA is equal to GB (3), DA equal to DB (4), and the side (3) Hypoth. GD common to both, the angles GDA and GDB are equal (5), and therefore are right angles (6); but the angle CDB is a right angle (4), therefore GDB is equal to CDB (7), a part equal to the whole, which is absurd: B.1. G therefore is not the centre of the circle ACB; and in B.1. the same manner it can be proved that no other point without the line CE is the centre; therefore the centre is in the line CE, and therefore is the point F.

Schol. Hence it is evident that, if any line terminated in a circle be bisected by a perpendicular, that perpendicular, if produced, will pass through the

centre.

(6) Def. 11.

(7) Ax. 11.

PROP. II. THEOR.

If any two points (A and B) be taken in the circum- Fig. 6. ference of a circle, the right line which joins them falls See N. within the circle.

For, if it be possible, let AEB be a right line, in which the point È is without the circle; and draw DA, DE, and DB.

(1) Def. 15.

(2) Prop. 5. B. 1.

Because in the triangle ADB, the sides DA and DB are equal (1), the angle DBA is equal to DAB (2); but the external angle DEA is greater than the internal angle DBA (3), therefore greater than the angle DAB, and therefore the side DA is greater than the side DE (4); but the right line DF is equal to DA (1), and Prop.19.

(3) Prop.16.

B.1.

(4)

B. 1.

therefore is greater than DE, a part greater than the whole, which is absurd; therefore the line AEB is not a right line; and in the same manner it can be demonstrated that if the point E be in the circumference, the line is not a right line.

Therefore every point of a right line falls within the circle.

Fig. 7.

(1) Def. 15. B. 1.

(2) Hypoth.

(3) Prop. 8.

B. 1.

(4) Def. 11.

B. 1.

B. 1.

PROP. III. THEOR.

If a right line (BI), drawn through the centre of a circle, bisect a right line (CF), which does not pass through the centre, it is perpendicular to it.

And if it cut it at right angles, it bisects it.

Part 1. Draw AC and AF; in the triangles AOC, AOF, the side AC is equal to AF (1), and also OC to OF (2), and AO is common to both; therefore the angle AOC is equal to AOF (3), therefore each of them is a right angle (4), and therefore BO is perpendicular to CF (4).

Part 2. Because the triangle FAC is isosceles (1), the (5) Def. 15. angle AFC is equal to the angle ACF (5); therefore in the triangles CAO, FAO, the angles ACO and AFO are equal, also AOC and AOF are equal (2), and the side AO opposite to the equal angles ACO, AFO, is common (6) Prop.26, to both; therefore the side OC is equal to OF (6), and therefore the right line CF is bisected.

B. 1.

Fig. 8.

PROP. IV. THEOR.

If in a circle two right lines cut one another, and do not both pass through the centre, they do not bisect one another.

If one of the lines pass through the centre, it is evident that it cannot be bisected by the other, which does not pass through the centre.

But if neither of the lines BC and FL pass through the centre, draw OA from the centre to their intersec