square

B D

First, let AD fall within the triangle ABC. Then because the straight line CB is divided in D, the squares of CB, BD are equal to twice the rectangle contained by CB, BD, and the of DC; to each add the square of AD; therefore the squares of CB, BD, DA, are equal to twice the rectangle CB, BD, and the squares of AD, DC; but the square of AB is equal to the squares of BD, DA, because the angle BDA is a right angle; and the square of AC is equal to the squares of AD, DC; therefore the squares CB, BA are equal to the square of AC, and twice the rectangle CB, DB; that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB, BD. Next, let AD fall outside the triangle ABC. Then, because D is a right angle, the angle ACB is greater than a right angle; and the square of AB is equal to the squares of AC, CB, and twice the rectangle

BC, CD; to each add the square of BC; therefore the squares of AB, BC are equal to the square of AC, twice the square of BC, and twice the rectangle BC, CD; but because BD is divided in C, the rectangle DB, BC is equal to the rectangle BC, CD, and the square of BC; and twice the rectangle DB, BC is equal to the rectangle BC, CD, and twice the square of BC: therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: therefore the square of AC is less than the` squares of AB, BC, by twice the rectangle DB, BC. Lastly, let AC be perpendicular to BC. Then BC is the straight line between the perpendicular and the acute angle at B; and the squares AB, BC, are equal to the of AC, and twice the square of

square BC.

B

XIV.—To describe a square equal to a given rectilined?

figure.

Let A be the given rectilineal figure. Describe the rectangular parallelogram BCDE equal to A. Then, if the sides of A it are equal to one another, it is a square. But if not,

B

G E

C

D

produce BE to F, and make EF equal to ED, bisect BF in G; from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to meet the circumference in H. The square described upon EH shall be equal to the given rectilineal figure A. Join GH. Because BF is bisected in G, and divided into two unequal parts in E; the rectangle BE, EF, together with the square of EG, are equal to the square of GF; but GF is equal to GH; therefore the rectangle BE, EF, together with the square of EG, is equal to the square of GH; but the squares of HE, EG are equal to the square of GH; therefore the rectangle BE, EF, together with the square of EG, are equal to the squares of EG, EH; take away the common square of EG; therefore the rectangle BE, EF is equal to the square of HE. But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH; but BD is equal to the rectilineal figure A; therefore the square of EH is equal to the rectilineal figure A.

GEOMETRY.-BOOK III:

DEFINITIONS.

I. Equal circles have equal diameters, or the straight lines from the centres to the circumferences are equal. II. A straight line touches

a circle when it meets it, and being produced does not cut

it.

III. Circles touch one another, which meet but do not cut one another.

IV. Straight lines are equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal.

V. The straight line on which the greater perpendicular falls is farther from the centre.

VI. A segment of a circle is contained by a straight line and the circumference it cuts off.

VII. The angle of a segment is that contained by the straight line and the circumference.

VIII. An angle in a segment is that contained by two straight lines drawn from any point in the circumference of the segment to the extremities of the straight line which is the base of the segment.

D

IX. And an angle stands on the circumference intercepted between the straight lines which contain the angle.

X. A sector is contained by two straight lines drawn from the centre and the circumference.

XI. Similar segments have equal angles.

PROPOSITION I.

To find the centre of a given circle. Draw within ABC any straight line AB, anu bisect AB at D.

From D draw DC at right angles to AB; produce CD to meet the circumference at E, and bisect CE at F. F is the centre of ABC.

If F be not the centre, if possible, let G be the centre; and join GA,

GD, GB. Because DA is equal to DB, and DG common to ADG, BDG; AD, DG are equal to BD, DG, each to each; and GA is equal to GB, because they are drawn from the centre G; therefore ADG is equal to BDG. Therefore the angle BDG is a right angle; but BDF is also a right angle. Therefore BDG is equal to BDF, the less to the greater, which is impossible. Therefore G is not the centre.

In the same manner it may be shewn that no other point out of CE is the centre; and since CE is bisected at F, any other point in CE divides it into unequal parts, and cannot be the centre. Therefore no point but F is the centre; that is, F is the centre.

COROLLARY.-From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the straight line bisecting the other.

II-If any two points be taken in the circumference of a circle, the straight line which joins them shall full within the circle.

The straight line from A to B falls within the circle ABC.

A

C

D

E

B

If not, let it fall, if possible, without, as AEB. Find D the centre of the circle ABC; and join DA, DB; in the arc AB take any point F, join DF, and produce it to meet AB at E. Because DA is equal to DB, the angle DAB is equal to DBA. And because AE is produced to B, the exterior angle DEB is greater than the interior opposite angle DAE. But DAE is equal to DBE; therefore DEB is greater than DBE. But the greater angle is subtended by the greater side; therefore DB is greater than DE. But DB is equal to DF; therefore DF is greater than DE, the less than the greater; which is impossible. Therefore the straight line from A to B does not fall without the circle. the same manner it may be shewn that it does not fall on the circumference. Therefore it falls within the circle.

In

III.-If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it cuts it at right angles; and if it cut it at right angles it bisects it.

Let CD drawn through the centre of ABC bisect AB, which does not pass through the centre in F; CD cuts AB at right angles.

Take E the centre of the circle, and join EA, EB. Because AF is equal to FB, and FE common to AFE, BFE; AF, FE are equal to BF, FE, each to each; and EA is equal to EB; therefore the angle AFE is equal to BFE. Therefore each of the angies AFE, CFE is a right angle.

Fl

D

Therefore CD, drawn through