Book IV. circumference of the circle touches each angle of the figure about which it is circumfcribed. 7. A straight line is faid to be applied or placed in a circle, when the extremities of it are in the circumference of the circle. PROP. I. In a given circle to apply a straight line, equal to a given straight line, which is not greater than the diameter. A Let the given circle be ABC; and D the given straight line, not greater than the diameter of the circle: it is required to apply in the circle ABC a straight line equal to the straight line D. Let BC the diameter of the circle ABC be drawn: if therefore BC be equal to D, the thing required has been done for the ftraight line BC has been applied, in the circle ABC, equal to the ftraight line D. But if not, BC is greater than D (by supp.); and (by 3. 1.) make CE equal to D ; and with the center C, and at the distance CE let the circle AEF be described; and let CA be joined. Wherefore because the point C is the center of the circle AEF CA is equal to CE; but D is equal to CE; therefore alfo D is equal to CA. B E F D Wherefore in the given circle ABC, the ftraight line AC has been applied, equal to the given straight line D, which is not greater than the diameter of the circle. Which was to be done. In a given circle, to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle; it is required to infcribe in the circle ABC a triangle equiangular to the triangle DEF. 1 Let GAH be drawn, touching the circle ABC in the point A ; Book IV. and let the angle HAC be made with the straight line AH, and at the point A in it, equal to the angle DEF: and the angle GAB, with the straight line AG, and at the point A in it, equal to the angle FDE; and let BC be joined. Therefore because a certain ftraight line HAG touches the circle ABC, and a certain straight line AC hath been drawn from the contact cutting it; therefore (by 32. 3.) the angle HAC is equal to the angle ABC, the angle in the alternate fegment of the B G A H circle: but the angle HAC is equal to the angle DEF; therefore the angle ABC is equal to the angle FED. Certainly for the fame reason also the angle ACB is equal to FDE; and therefore (by 32. 1.) the remaining angle BAC is equal to the remaining angle EFD : wherefore the triangle ABC is equiangular to the triangle DEF; and it has been infcribed in the circle ABC (by def. 3.4.). Wherefore a triangle equiangular to the given triangle hath been inscribed in the given circle. Which was to be done. To circumfcribe a triangle, about a given circle, equiangular to a given triangle. 1 Let ABC be the given circle, and DEF the given triangle; it is required to circumfcribe a triangle about the circle ABC equiangular to the triangle DEF, Let EF be produced towards both pa F be produced towards both parts, to the points H, G ; and let K the center of the circle ABC be taken; and let the ftraight line KB be drawn as it may happen and let the angle BKA be made with the straight line KB and at the point K in it equal to the angle DEG; and BKC equal to the angle DFH; and through the points A, B, C let the ftraight lines LAM, MBN, NCL be drawn touching the circle ABC. VOL. I. and L G H E F A C K M B N Book IV. And becaufe LM, MN, NL touch the circle ABC in the points, A, B, C; and from the center K; KA, KB, KC have been drawn to the points A, B, C; therefore (by 18. 3.) the angles at the points A, B, Care right angles. And because the four angles of the quadrilateral figure AMBK are equal to four right angles (by 32. 1.); [fince the quadrilateral figure AMBK is divifible into twe triangles] of which the angles KAM, KBM are two right angles; wherefore the remaining angles AKB, AMB are equal to two right angles; but the angles DEG, DEF are also (by 13.11.) equal to two right angles; wherefore the angles AKB, AMB are equal to the angles DEG, DEF; of which AKB is (by conft.) equal to DEG; therefore the remaining angle AMB is equal to the remaining angle DEF. Certainly in the fame manner it will be demonstrated that the angle LNM is equal to the angle DFE therefore the remaining angle MLN is equal to the remaining angle EDF (by 32. 1.); therefore the triangle LMN is equiangular to the triangle DEF and (by def. 4. 4.) it is circumfcribed about the circle ABC. Wherefore a triangle, has been circumfcribed about the given circle, equiangular to the given triangle. Which was to be done. Let ABC be the given triangle; it is required to infcribe a circle in the triangle ABC. Let the angles ABC, BCA be cut in halves by the straight lines BD, CD; and let them meet one another in the point D; and from the point D let DE, DF, DG be drawn perpendiculars to the ftraight lines AB, BC, CA. t And because the angle ABD is equal to the angle CBD, for ABC is cut in halves; alfo the right angle BED is equal to the A right angle BFD; therefore there are two F on account of the Wherefore the circle EFG is infcribed in the given triangle ABC. Which was to be done. To circumfcribe a circle about a given triangle. Let ABC be the given triangle: it is required to circumscribe a circle about the given triangle ABC. Let AB, AC be cut in halves in the points D and E; and let DF and EF be drawn from the points D, E at right angles to AB, AC; they will meet either within the triangle ABC or in the straight line BC or without the triangle ABC. : First let them meet within at the point F, and let BF, FC, FA be joined and because AD is equal to DB; and DF common and at right angles; therefore (by 4. 1.) the bafe AF is equal to the bafe FB: Certainly in like manner we fhall demonftrate that CF is. alfo equal to FA; fo that (by com. not..) BF is also equal to 0 2 FC; Book IV. Book IV. FC; therefore the three straight lines FA, FB, FC are equal to one another; wherefore a circle defcribed with the center F and at the distance of any one of the lines FA, FB, FC will also pafs through the remaining points; and the circle will be circumscribed about the triangle ABC; and let it be described as the circle. ABC. But let DF, EF meet in the ftraight line BC, as it is in the second figure; and let AF be joined: Certainly in the fame manner we shall demonftrate that the point F is the center of the circle circumfcribed about the triangle ABC. But let DF, EF meet without the triangle ABC, again in the point F; as it is in the third figure; and let AF, FB, FC be joined and again because AD is equal to DB, and DF common and at right angles; therefore (by 4. 1.) the bafe AF is equal to the base FB: Certainly in the fame manner we fhall demonftrate that CF is equal to FA; fo that alfo BF is equal to FC; therefore again the circle described with the center Fand at the distance of any one of the lines FA, FB, FC will also pass through the remaining points; and will be circumfcribed about the triangle ABC; and let it be deferibed as ABC. Wherefore a circle has been circumfcribed about the given triangle. Which was to be done. !, mir Cor. And it is manifest that, when the center of the circle falls within the triangle, the angle BAC being in a fegment greater than a femicircle, is lefs than a right angle; but when it falls in BC being in a femicircle, it will be a right angle; but when the center falls without the triangle ABC, the angle BAC being in a fegment less than a femicircle, is greater than a right angle. So that also when the given triangle is acute angled the ftraight lines DF, EF will meet within the triangle; but when BAC is a right angle; they will meet in BC;.. but when it is greater than a right angle, without the triangle ABC. 1. PROP. |