therefore a right angle; consequently ACBF is a semicircle (III. 19. cor.) and AF the diameter. The angle ADF in the opposite semicircle is hence a right angle (III. 19.), and the square of the diameter AF is equal to the squares of AD and FD, or to the sum of the squares of the four segments EA, EB, ED, and EC intercepted between the circumference and the point E.

PROP. XXVI. THEOR.

If through a point, within or without a circle, two straight lines be drawn to cut the circumference; the rectangle under the segments of the one, is equivalent to that contained by the segments of the other.

Let the two straight lines AD and AF be extended through the point A, to cut the circumference BFD of a circle; the rectangle contained by the segments AE and AF of the one, is equivalent to the rectangle under AB and AD, the distances intercepted from A in the other.

For draw AC to the centre, and produce it both ways to terminate in the circumference at G and H; let fall the perpendicular CI upon BD (I. 6.), and join CD.

Because CI is perpendicular to AD, the difference between the squares of CA and CD, the sides of the triangle ACD is equivalent to the difference between the squares of the segments AI and ID the segments of the base (II. 21. cor.); and the difference between the squares of two straight lines being equivalent to the rectangle under their sum and their

B

G

E

H

I

E

F

difference (II. 17.), the rectangle contained by the sum and difference of AC, CD is equivalent to the rectangle contained by the sum and difference of AI, ID. But since the radius CG is equal to CH, the sum of AC and CD is AH, and their difference is AG; and because the perpendicular CI bisects the chord BD (III. 4.), the sum of AI and ID is AD, and their difference AB. Wherefore the rectangle AH, AG is equivalent to the rectangle AB, AD. In the same way it is proved, that the rectangle AH, AG is equivalent to the rectangle AE, AF; and consequently the rectangle AE, AF is equivalent to the rectangle AB, AD.

Cor.. 1. If the vertex A of the straight lines lie within

the circle and the point I coincide with

it, BD, being then at right angles to CA, is bisected at A (III. 4.), and the rectangle AB, AD is the same as the square of AB. Consequently the square of a perpendicular AB limited by the circum

ference is equivalent to the rectangle under the segments AG, AH of the diameter.

Cor. 2. If the vertex A lie without the circle and the

point I coincide with B or D, the an

B

H

gle ABC being then a right angle, the incident line AB must be a tangent (III. 20.), and consequently the two points of section B and D must coalesce in a single point of contact. Wherefore the rectangle under the distances AB, AD becomes the same as the square of AB; and consequently

H

the rectangle contained by the segments AG, AH of the diameter, is equivalent to the square of the tangent AB.

PROP. XXVII. PROB.

To construct a square equivalent to a given rectilineal figure.

Let the rectilineal figure be reduced by Proposition 6. Book II. to an equivalent rectangle, of which A and B are the two containing sides; draw an indefinite straight

A

B

F

line CE, in which take the part CD

equal to A and DE to B, on C describe a semicircle, and erect the per

D

E

pendicular DF from the diameter to meet the circumference: DF is the side of the square equivalent to the given rectilineal figure.

For, by Cor. 1. to the last Proposition, the square of the perpendicular DF is equivalent to the rectangle under the segments CD, DE of the diameter, and is consequently equivalent to the rectangle contained by the sides A and B of a rectangle that was made equivalent to the rectilineal figure.

PROP. XXVIII. THEOR.

A quadrilateral figure may have a circle described about it, if the rectangles under the segments made by the intersection of its diagonals be equi

valent, or if those rectangles are equivalent which are contained by the external segments formed by producing its opposite sides.

Let ABCD be a quadrilateral figure, of which AC and BD are the diagonals, and such that the rectangle AE, EC is equivalent to the rectangle BE, ED; a circle may be made to pass through the four points A, B, C, and D. For describe a circle through

the three points A, B, C (III. 9. cor.), and let it cut BD in G. Because AC and BG intersect each other within a circle, the rectangle AE, EC is equivalent to the rectangle BE, EG (III. 26.); but

F

B

E

the rectangle AE, EC is by hypothesis equivalent to the rectangle BE, ED. Wherefore BE, EG is equivalent to BE, ED; and these rectangles have a common base BE, consequently (II. 3. cor.) their altitudes EG and ED are equal, and hence the point G is the same as D, or the circle passes through all the four points A, B, C, and D.'

Again, if the opposite sides CB and DA be produced to meet at F, and the rectangle CF, FB be equal to DF, FA, a circle may be described about the figure.

For, as before, let a circle pass through the three points A, B, C, but cut AD in H. And from the property of the circle, the rectangle CF, FB is equivalent to HF, FA; but the rectangle CF, FB is also equivalent to DF, FA; whence the rectangle HF, FA is equivalent to DF, FA, and the base HF equal to DF, or the point H is the same as D.