PROP. X. THEO R.

If a right line cutting a circle makes an angle at one end of the common bafe of two fegments of that circle equal to the angle in either fegment, and any other right line be drawn from the other end of that bafe to the first mentioned right line, the part of this last line within the circle, the common bafe of the fegments, and the line itself, will be proportionals.

Let the right line AC be the common base of the fegments ABC, AFC of a circle, and let the right line E CD cutting the circle, make the angle ACD equal to the angle A B C in the fegment A B C, or the angle ACE equal to the angle AFC in the fegment AFC, and let any right line A D, cutting the circle in B, or any right line AE cutting the circle in F, be drawn from the end A of the common base AC of the fegments, to the line ED: I say the right lines A B, AC, AD, as alfo A F, AC, AE will be proportionals.

For join CB, CF.

Then because [by conftruction] the angle A CD is equal to the angle ABC, and fince [by 21. 3.] the angle ABC

E

B

is always of the fame magniDtude however the line A D be drawn. And because the angle A is common to both the triangles ABC, ACD; therefore [by 32. 1.] the angles CD B, ACB will be equal; and fo the triangles ABC, ACD will be always equiangular. Therefore [by 4. 6.] as AB is to AC, fo will AC be to A D. In like manner the triangles ACE, ACF will be always equiangular. Wherefore [by 4.

6. as AF is to A c, fo will A c be to A E.
Therefore, &c. Which was to be demonftrated.

Coral

Corollary. Hence if any right line a D, drawn from A, cuts the circle in B, the rectangle under AD and A B will be always of the fame magnitude, viz, equal to the fquare of A c.

PROP. XI.

If A c be the diameter of a circle, which the right line D E cuts at right angles; and from the end a of the diameter the right line AB be drawn meeting the circumference in B, and the right line DE in F. I fay the three right lines AB, AD, AF will be proportionals.

A

For first let the interfection F fall within the circle, and draw B D. Then because the arches E AE, AD are equal, the anglés at the circumference EDA, A BD ftanding upon them will be equal. And fo in the triangles A B D, ADF the angles A B D, ADF [by

21. 3.] are equal. But the angle at A is common to both triangles. Wherefore [by 32. 1.] the remaining angles ADB AFD will be equal to one another. And fo the triangles A B D, ADF will be equiangular. Wherefore [by 4. 6.] as BA is to A D, fo will AD be to AF. But now let the point of interfection fall without the circle, and draw bH parallel to D E, meeting the right line AD in H. Then from what has been already demonftrated, it will be as DA is to Ab, fo is ab to AH, the triangles AbD, Aнb being equiangular; that is [by conftruction and 2. 6.] fo will af be to AD, therefore Af, AD, Ab will again be proportionals,

Therefore, &c. Which was to be demonstrated.

If a right line cutting a circle makes an angle at one end of the common base of two fegments of that circle equal to the angle in either fegment, and any other right line be drawn from the other end of that bafe to the first mentioned right line, the part of this last line within the circle, the common bafe of the fegments, and the line itself, will be proportionals.

Let the right line AC be the common base of the fegments A BC, AFC of a circle, and let the right line ECD cutting the circle, make the angle ACD equal to the angle ABC in the fegment A B C, or the angle ACE equal to the angle AFC in the fegment AF C, and let any right line A D, cutting the circle in B, or any right line A E cutting the circle in F, be drawn from the end A of the common base AC of the fegments, to the line ED: I say the right lines AB, AC, AD, as also a F, AC, AE will be proportionals.

For join CB, CF.

Then because [by conftruction] the angle ACD is equal to the angle ABC, and fince [by 21. 3.] the angle ABC

E

B

is always of the fame magniDtude however the line A D be drawn. And because the angle A is common to both the triangles ABC, ACD; therefore [by 32. 1.] the angles CDB, A CB will be equal; and fo the triangles ABC, ACD will be always equiangular. Therefore [by 4. 6.] as AB is to AC, fo will AC be to A D. In like manner the triangles ACE, ACF will be always equiangular. Wherefore [by 4.

6.] as AF is to A c, fo will A c be to A E. Therefore, &c. Which was to be demonstrated.

Coral

Corollary. Hence if any right line A D, drawn from a, cuts the circle in B, the rectangle under AD and A B will be always of the fame magnitude, viz, equal to the fquare of A C.

PROP. XI.

If cbe the diameter of a circle, which the right line D E cuts at right angles; and from the end a of the diameter the right line AB be drawn meeting the circumference in B, and the right line DE in F. I fay the three right lines A B, AD, AF will be proportionals.

For firft let the interfection F fall within the

circle, and draw B D. Then because the arches E AE, AD are equal, the angles at the circumference EDA, ABD ftanding upon them will be equal. And fo in the triangles A B D, ADF the angles A B D, ADF [by

21. 3.] are equal. But the angle at A is common to both triangles. Wherefore [by 32. 1.] the remaining angles ADB, AFD will be equal to one another. And fo the triangles A B D, ADF will be equiangular. Wherefore [by 4. 6.] as BA is to A D, fo will AD be to AF. But now let the point of interfection fall without the circle, and draw bH parallel to D E, meeting the right line AD in H. Then from what has been already demonftrated, it will be as DA is to Ab, fo is ab to AH, the triangles AbD, Aнb being equiangular; that is [by conftruction and 2. 6.] fo will af be to AD, therefore Af, AD, Ab will again be proportionals,

Therefore, &c. Which was to be demonftrated.

1

PROP.

PROP. XII. THEOR.

If any triangle be infcribed in a circle, and a perpendicular be let fall from the vertical angle upon the bafe, the rectangle under the fides of the triangle will be equal to the rectangle under that perpendicular and the diameter of the circumfcribed circle.

Let ADB be a triangle, whose altitude is DG, and let a circle, whose centre is C, circumfcribe it. Draw the diameter DH, and join HR: Ifay the rectangle under the fides AD, DB will be equal to the rectangle under the perpendicular DG, and the diameter DH of the circumfcribed circle.

For because the angles A, H [by. 21. 3.] are equal,,as standing upon the fame right line DB, and the angles

D

A

H

B

"G, DEH right angles [by fuppofition, and 31. 3. the triangles ADG, DBH will be fimilar. And therefore as A D is to DG, fo will DH be to D B. Therefore [by 16. 6.] -the rectangle under the fides A D, DB of the infcribed triangle will be equal to the rectangle under the perpendicu lar DG and the diameter DH of the circumfcribing circle. Therefore, &c. Which was to be demonftrated.

PROP. XIII. THEOR.

If any trapezium be infcribed in a circle, the rectangle under the diagonals will be equal to the fum of the rectangles under each of the oppofite fides.

Let there be a trapezium A B C D infcribed in a circle, whofe diagonals are AC, BD. I fay the rectangle under

AC, BD