middle point E of the semi-circumference AEB; the squares of CD and CE together will be double of the square of the semi-diameter.

Join DO, OE. The angle EOC is a right angle, and the square of EC is equal to the sum of the squares of EO and OC; but the square of DC is equal to the difference of the squares of DO and OC; .. the squares of EC and CD together are equal to the squares of EO and DO together, i. e. are double of the square of EO.

(5.) If a straight line be drawn at right angles to the diameter of a circle, and be cut by any other line ; the rectangle contained by the segments of this cutting line, together with the square of that part of the perpendicular line which is intercepted between it and the diameter, is always of the same magnitude.

Let AB be drawn at right angles to CD the diameter of the circle ABC; and let it be cut in G, by any other line EF; the rectangle EG, GF, together with the square of HG is of invariable magnitude.

CE

A

B

G

D

T

For the rectangle EG, GF is equal to the rectangle AG, GB, and the rectangle AG, GB together with the square of HG is equal to the square of AH, .. the rectangle EG, GF together with the square of HG is equal to the square of half AB, which is always the same.

LL

(6.) A straight line being drawn from the centre of a quadrant bisecting the arc and meeting a tangent drawn from one extremity; if from any point in the bounding radius a line be drawn parallel to the tangent, the sum of the squares of the segments of it, cut off by the aforesaid line and by the circumference will be equal to the square of the radius.

C

I

From the centre O let OC be drawn bisecting the quadrantal arc AB, and meeting a tangent to the point A in C. From any point D in 40 draw a perpendicular DE; the squares of DF and DE are together equal to the square of OB.

R

Join FO. Since the angle DOE is half a right angle, and the angle at D a right angle, .. DEO is half a right angle, and equal to DOE; whence DE=DO. Now the squares of DO and DF are together equal to the square of OF; .. also the squares of DE and DF are together equal to the square of OF, or OB. In the same manner it may be shewn that the squares of GH and GI are together equal to the square of OB.

(7.) If from a point without a circle there be drawn two straight lines, one of which is perpendicular to a diameter, and the other cuts the circle; the square of the perpendicular is equal to the rectangle contained by the whole cutting line and the part without the circle, together with the rectangle contained by the segments of the diameter.

From the point A let AB be drawn perpendicular

to CD the diameter of the circle DEC, and AFE cutting the circle; the square of AB is equal to the rectangles EA, AF, and CB, BD together.

Through the centre O draw AGH. The squares of AB, BO are equal to the square

of AO, i.e. to the rectangle HA, AG together with the square of GO (Eucl. ii. 6.), i. e. to the rectangle HA, AG together with the rectangle DB, BC and the square of OB; and ... the square of AB is equal to the rectangles HA, AG and DB, BC together.

(8.) If any straight line be drawn perpendicular to the diameter of a given circle, and produced to cut any chord; the rectangle contained by the segments of the diameter will be less or greater than the rectangle contained by the segments of the chord, by the square of the line intercepted between them, according as it is drawn without or within the circle.

Let AB meet the diameter CD of the circle CGD at right angles in the point E, and any other chord GH in F; the rectangle CE, ED is less or greater than the rectangle GF, FH, by the square of EF, according as AB is without or within the circle.

K

G

E

B

D

E

H

Take O the centre of the circle, and through it draw FOK cutting the circle in I and K. Then because KI is bisected in 0, and produced to F, the rectangle KF, FI together with the square of OI is equal to the square of OF, i. e. to the squares of OE and EF. But when E

is without the circle, the rectangle CE, ED together with the square of OD is equal to the square of OE, ..the rectangle KF, FI together with the square of OI is equal to the rectangle CE, ED, together with the squares of OD, and EF. And since OI is equal to OD, and the rectangle KF, FI is equal to the rectangle GF, FH, (Eucl. iii. 36. Cor.); .. the rectangle GF, FH is equal to the rectangle CE, ED together with the square of EF. In nearly the same manner it is demonstrated if AB be within the circle.

(9.) If a diameter of a circle be produced to bisect a line at right angles, the length of which is the double of a mean proportional between the whole line through the centre and the part without the circle; and from any point in the double of the mean proportional a line be drawn cutting the circle; the sum of the squares of the segments of the double mean proportional will be equal to twice the rectangle contained by this cutting line and the part without the circle.

H R

A

Let the diameter BA produced bisect DCE at right angles, and let CD and CE be each a mean proportional between AC and CB; and through any point F let FGH be drawn cutting the circle in G and H; the squares of DF, and FE are together equal to twice the rectangle GF, FH.

D

F

C

E

Since the rectangle AC, CB is equal to the square of CD, the rectangle AC, CB together with the square of CF is equal to the squares of CD and CF. But (viii. 8.)

the rectangle AC, CB together with the square of CF is equal to the rectangle GF, FH; whence the rectangle GF, FH is equal to the squares of CD and CF together; and the doubles of equals are equal, .. twice the rectangle GF, FH is equal to twice the squares of CD and CF together, i. e. to the squares of DF and FE together (Eucl. ii. 9.).

COR. If from F tangents be drawn to the circle, the sum of their squares will together be equal to the sum of the squares of DF and FE.

(10.) If from a point without a circle two straight lines be drawn, one through the centre to the circumference, and the other perpendicular to it, and on the former a mean proportional be taken between the whole line and the part without the circle; any other line passing through that extremity of the mean proportional which is within the circle, and terminated by the circumference and perpendicular, will be similarly divided.

From a point C without the circle ABG, let CAB be drawn through the centre; take a point D such that AC: CD CD: CB; and from Clet CE

Er

F

B

be drawn perpendicular to CB; if through D, any line EFG be drawn terminated by the circumference and the perpendicular CE, EF ED ED: EG.

For the rectangle AC, CB together with the square of CE is equal, by construction, to the squares of DC, CE, i. e. to the square of DE. But (viii. 8.) the rectangle AC, CB together with the square of CE is equal