a 32. 3. Deductions. 1. Upon a given finite right line to describe the segment of a circle, which shall be similar to a given segment. 2. Given the base, the verticle angle, and the difference of the other two sides, to construct the triangle. 3. The base, the vertical angle, and the altitude, being given to construct the triangle. PROPOSITION XXXIV. PROBLEM. From a given circle to cut off a segment which shall contain an angle equal to a given rectilineal angle. Let ABC be a given circle, and D the given rectilineal angle. It is required from the circle ABC to cut off a segment which shall contain an angle equal to the angle D. Draw the right line EF, touching the circle ABC, in the point в, and at the right line BF, and at the point в in it make the angle FBC equal to the angle D. Therefore because a certain right line EF touches the circle ACB in the point в, and from the point of contact, BC is drawn, the angle FBC will be equal to that in the alter nate segment of the circle BAC. But the angle FBC is equal to the angle at D; wherefore also the angle in the segment BAC will be equal to the angle at D. Therefore from a given circle ABC, a segment BAC is cut off containing an angle equal to the given rectilineal angle at If in a circle two right lines mutually cut one another, the rectangle contained under the segments of one of them is equal to the rectangle contained under the segments of the other. For in the circle ABCD let the two right lines AC, BD, mutually cut one another in the point E. The rectangle A B D F H E B a contained under AE, EC, is equal to that contained under DE, EB. If AC, BD, pass through the centre so that E be the centre of the circle ABCD, it is manifest the right lines AE, EC, DE, EB, being equal, the rectangle contained under AB, EC, is equal to that which is contained under DE, EB. If AC, DB, do not pass through the centre, find the centre of the circle ABCD, which let be F, and from F draw FG, FH, perpendicular to AC, BD, and join FB, FC, FE. Because therefore a certain right line GF drawn through the centre cuts the right line AC not drawn through the centre at right angles, it shall bisect it; wherefore AG 3. 3. is equal to AC, and because the right line AC is divided into equal parts at the point G, and into unequal at the point E, the rectangle contained under AE, EC, together with the square of EG, will be equal to the square of GC, add the common square of GF. Wherefore the 5. 2. rectangle contained under AE, EC, together with the squares of EG, GF, is equal to the squares CG, GF, but the square of FE is equal to the squares of EG, GF; also 47. 1. the square described upon Fc is equal to the squares of CG, GF. Therefore the rectangle under AE, EC, together with the square of FE, is equal to the square of FC. But CF is equal to FB; wherefore the rectangle under AE, EC, together with the square of EF, is equal to the square described upon FB. For the same reason, the rectangle under DE, EB, together with the square of FE, is equal to the square of FB. But it was shown that the rectangle under AE, EC, together with the square of FE, is equal to the square of FB. Wherefore the rectangle under AE, EC, together with the square of FE, is equal to the rectangle under DE, EB, together with the square of FE; take away the common square of FE; therefore the remaining rectangle under DE, EC, will be equal to the remaining rectangle under DE, EB. Wherefore if in a circle, &c. Q. E. D. Deduction. с To make a rectangle which shall be equal to a given square, and shall have its two adjacent sides, together, equal to a given right line, the side of the given square being less than half of the given right line. a 47.1. • 5. 2. PROPOSITION XXXVI. THEOREM. If any point be taken without a circle, and from it two right lines be let fall on the circle, one of which cuts the circle, and the other touches it, the rectangle which is contained by the whole cutting line, and that part between the point taken without the circle, and the convex circumference of the circle, will be equal to the square of the touching line. D D For without the circle ABC take any point D, and from it let fall the two right lines DCA, DB, to the said circle, and let DCA cut the circle ABC, but DB touch it. The rectangle contained under AD, DC, is equal to the square B of DB. For DCA either passes through the centre, or it does not. First, let DA pass through the centre of the circle ABC, A E. B E which let be E, and join EB. The angle EBD will be a right angle, because the right line AC is bisected in E, and CD is added to it; the rectangle under AD, DC, together with the square of EC, will be equal to the square of ED, but CE is equal to EB; wherefore the rectangle under AD, DC, together with the square of EB, will be equal to the square of ED, but the square of ED is equal to the squares of EB, BD, for EBD is a right angle. Take away the common square of EB; wherefore the remaining rectangle under AD, DC, will be equal to the square of DB. Secondly, let DCA not pass through the centre of the circle ABC, and find the centre E, and draw EF perpendicular to AC, and join EB, EC, ED. Therefore EFD is a right angle. And because a certain right line EF drawn through the centre cuts the right line AC not drawn through the centre at right angles, it shall also bisect it; wherefore AF is equal to FC. Again, because the right line AC is bisected in F, and CD is added to it, the rectangle under AD, DC, together with the square of FC, is equal to the square of FD; add the common square of FE; therefore the rectangle under AD, DC, together with the squares of FC, FE, is equal to the squares of DF, FE. But the square of DE is equal to the squares of DF, FE, because a EFD is a right angle, but the square of CE is equal to the COROLLARIES. 1. (Clavius.) From this 36th proposition, it is manifest if, from any point without a circle, several right lines are drawn cutting the circle, the rectangles contained under the whole lines and the parts without the circle, are equal to one another. 2. (Clavius.) It is also proved that two right lines drawn from the same point which touch the circle are equal to one another. 3. (Clavius.) It is also evident from the same point only two right lines can be drawn which can touch the circle. PROPOSITION XXXVII. THEOREM. If any point be taken without a circle, and from it two right lines be let fall to the circle, one of which cuts the circle, and the other falls upon it; also let the rectangle contained by the whole line cutting the circle, and the part between the point taken without the circumference, and the convex circumference be equal to the square of the line meeting the circle, the line which meets it shall touch the circle. For without the circle ABC take any point D, and from it let fall the two right lines DCA, DB. Let DCA cut the circle, and DB fall upon it, and let the rectangle AD, DC, be equal to the square of DB; then DB touches the circle ABC. For draw the right line DE touching the circle ABC, and find the centre of the circle ABC,a a 1. 3. which let be F; join FE, FB, FD; wherefore the angle FED is a right angle. And because DE touches the b 21.3. B D E circle ABC, but DCA cuts it, the rect- Deductions. b 1. To describe a circle which shall touch two given right lines and pass through a given point between them. 2. To describe a circle which shall pass through two given points, and touch a given right line, the given points being both on the same side of the right line. |