1

tained by BE and EF, because EF is equal to ED ; therefore BD is equal to the square of EH ; and BD is also equal to the rectilineal figure A ; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. Which was to be done.

PROP. A. THEOR. If one side of a triangle be bisected, the sum of the squares of the other

two sides is double of the square of half the side bisected, and of the square of the line drawn from the point of bisection to the opposite angle of the triangle.

Let ABC be a triangle, of which the side BC is bisected in D, and DA drawn to the opposite angle ; the squares of BA and AC are together double of the squares of BD and DA.

From A draw AE perpendicular to BC, and because BEA is a right angle, AB: = (47. 1.) BE + AE and AC2 =CE2 +AE ; wherefore AB? + i

А AC2 =BE +CE2+2 AE2. But because the line BC is cut equally in D, and unequally in E, BE2 + CE = (9.2.) 2BD: +2DE2 ; therefore AB+ AC2 2BD2 +2DE2,2AE2.

Now DE: + AE2 = (17. 1.) AD, and 2DE? + 2AE- = 2AD2 ; wherefore AB? + ACS = 2BD2 + ZAD2,

„B

E Therefore, &c. Q. E. D.

PROP. B. THEOR.

.

The sum of the squares of the diameters of any parallelogram is equal to

the sum of the squares of the sides of the parallelogram. Let ABCD be a parallelogram, of which the diameters are AC and BD; the sum of the squares of AC and BD is equal to the sum of the squares of AB, BC, CD, DA.

Let AC and BD intersect one another in E: and because the vertical angles AED, CEB are equal (15. 1.). and also the alternate angles EAD, ECB (29. 1.), the triangles ADE, CEB bave two angles in the one equal to two angles in the other, each to each ; but the sides AD and BC, which are oppo

А.

D site to equal angles in these triangles, are also equal (34. 1.); therefore the other sides

E which are opposite to the equal angles are also equal (26. 1.), viz. AE to EC, and ED to EB.

B B

Since, therefore, BD is bisected in E, ABP +AD:==(A. 2.) 2BE +2AE); and for the same reason, CD+BC2=2BE2 +2ECI=2BE3 +2AEbecause EC=AE. Therefore ABS +AD:+DC +BC:=

, = 4BE3 +4AE . But 4BE?=BD, and 4AE:=AC: (2. Cor. 8. 2.) because BD and AC are both bisected in E; therefore AB: + AD2 to CD + BC = BD2 + AC. Therefore the sum of the squares

&c. Q. E. D.

.

Cor. From this demonstration, it is manifest that the diameters of every parallelogram bisect one another.

OF

GEOMETRY.

BOOK III.

DEFINITIONS.

A.

HE radius of a circle is the straight line drawn from the centre to the circumference.

1. A straight line is said to touch

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

II.
Circles are said to touch one

another, which meet, but
do not cut one another.

III.
Straight lines are said to be equally dis-

tant from the centre of a circle, when
the perpendiculars drawnto them from
the centre are equal.

IV.
And the straight line on which the great-

er perpendicular falls, is said to be
farther from the centre

B.
An arch of a circle is any part of the circumference.

V.
A segment of a circle is the figure con-

tained by a straight line, and the arch
which it cuts off.

a

a

>

1

PROP. I. PROB.

To find the centre of a given circle.
Let ABC be the given circle ; it is required to find its centre.

Draw within it any straight line AB, and bisect (10. 1.) it in D; from the point D draw (11. 1.) DC at right angles to AB, and produce it to E, and bisect CE in F : the point F is the centre of the circle ABC.

For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each ; and the base

C GA is equal to the base GB, because they are radii of the same circle : therefore the angle ADG is equal (8. 1.) to the angle GDB : But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle (7. def. 1.) Therefore

АХ D the angle GDB is a right angle : But FDB is likewise a right angle ; wherefore the angle

E FDB is equal to the angle GDB,the greater

:

to the less, which is impossible : Therefore G is not the centre of the circle ABC: In the same manner, it can be shown, that no other point but F is the centre : that is, F is the centre of the circle ABC: Which was to be found.

a

Cor. 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 line which bisects the other.

PROP. II. THEOR.

If any two points be taken in the circumference of a circle, the straight

line which joins them shuli fali within the circle.

Let ABC be a circle, and A, B any two points in the circumference : the straight line drawn from A to B

с shall fall within the circle.

Take any point in AB as E; find D the centre of the circle ABC; join AD, DB and DE, and let DE meet the circumference in F: Then, because DA is equal to DB, the an

D gle DAB is equal (5. 1.) to the angle DBA; and because AE, a side of the triangle

E DAE, is produced to B, the angle DEB is

B greater (16. 1.) than the angle DAE; but

F DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE: Now to the greater angle the greater side is opposite (19. 1.); DB is therefore greater than DE: but BD is equal to DF ; wherefore DF is greater than DE, and the point E is therefore within the circle. be demonstrated of any other point between A and B, therefore AB is within the circle. Wherefore, if any two points, &c. Q. E. D.

The same may

PROP. III. THEOR.

If a straight line drawn through the centre of a circle bisect a straight

line in the circle, which does not pass through the centre, it will cut that line at right angles; and if it cut it at right angles, it will bisect it.

Let ABC be a circle, and let CD, a straight line drawn through the centre bisect any straight line AB, which does not pass through the centre, in the point F: It cuts it also at right angles.

Take (1. 3.) E the centre of the circle, and join EA, EB. Then because ÀF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other: