3. In an isosceles triangle, the perpendicular drawn from the vertex bisecting the base passes through the centre of the inscribed circle. 4. In a given circle, to inscribe three equal circles touching each other and the given circle.* PROPOSITION V. PROBLEM. To circumscribe a circle about a given triangle. Let ABC be the given triangle; it is required to circumscribe a circle about the given triangle ABC. 10. 1. Bisect the right lines AB, AC, in the points D, E, and b 11. 1. c 4. 1. from the points D, E, draw DF, FE, at right angles to AB, AC. They will meet either within the triangle ABC, or in the right line BC, or without BC. First, therefore, let them meet within the triangle in F, and join FB, FC, FA. And because AD is equal to BD, and the common side DF is at right angles; therefore the base AF is equal to FB. In like manner, we show also that CF is equal to AF, wherefore FB is also equal to FC; therefore the three FA, FB, FC, are equal to one another. Therefore with centre F, and distance any one of them FA, FB, FC, a circle described will pass through the remaining points, and the circle will be circumscribed 6 Def. 4. about the triangle ABC. Let it be circumscribed as ABC. But also, secondly, let DF, EF, meet in the right line BC in F, as it does in the second figure, and join In like manner it may be proved, that the point F is the centre of the circle circumscribed about the triangle ABC. AF.. But, thirdly, let DF, EF, meet without the triangle ABC in F, as in the third figure, and join AF, BF, CF. And because A D is equal to DB, also the common side DF *The student may find a very neat solution of this problem in Mr. Thomas Simpson's Algebra. is at right angles, therefore the base AF is equal to FB. 4. 1. In like manner we show also that FC is equal to FA, wherefore FB is equal to FC; therefore again with centre F, and distance any one of them, FA, FB, FC, a circle described will pass through the remaining points, and it will be circumscribed about the triangle ABC. Let it be described as ABC. Therefore a circle, &c. Q. E. F. COROLLARY. Hence it is manifest, when the centre of the circle falls within the triangle, the angle BAC, in a segment greater than a semicircle, is less than a right angle; but when the centre falls on the right line BC, the angle BAC, in a semircircle, is a right angle; and when the centre of the circle falls without the triangle BAC, in a segment less than a semicircle, is greater than a right angle. Wherefore, also, when the given angle is less than a right angle, AF, EF, will meet within the triangle, but when it is a right angle, on вC, and when a greater than a right angle, without BC. Deduction. In an equilateral triangle, the centre of the circumscribed circle coincides with the centre of the inscribed circle. PROPOSITION VI. PROBLEM. To inscribe a square in a given circle. Let ABCD be the given circle; it is required to inscribe a square in the circle ABCD. Draw the two diameters AC, BD, of the circle ABCD at right angles to one another, and join AB, BC, CD, DA. E And because BE is equal to ED, for E is the centre, and EA common, is at right angles; therefore the base AB is equal to the base AD;a and for the same reason BC, CD, are each equal to BA, AD; therefore the quadrilateral figure ABCD is equilateral. It is also rectangular. a 4. 1. For because the right line BD is a diameter of the circle ABCD, therefore BAD is a semicircle; consequently BAD is a right angle; for the same reason 31. 3. b a 17. 1. b 18.3. € 28.1. d 34. 1. each of the angles ABC, BCD, CDA, is a right angle; therefore the quadrilateral figure ABCD is rectangular; and it has been shown to be equilateral; wherefore it is a square, and is inscribed in the given circle ABCD. Therefore in a given circle, &c. Q. E. 1 Deductions. 1. The square inscribed in a circle is equal to twice the square of half the diameter. 2. To inscribe in a given circle a rectangle, which shall be equal to a given rectangle, whose diameter is equal to the diameter of the given circle. PROPOSITION VII. To circumscribe a square about a given circle. Let ABCD be a given circle; it is required to circumscribe a square about the circle ABCD. Draw the two diameters AC, BD of the circle ABCD at right angles to one another, and through the points A, B, C, D, draw FG, GH, HK, KF, touching the circle ABCD. a b G B E D C K Therefore because FG touches the H circle ABCD, and from E, the centre, draw EA to the point of contact A; therefore the angles at A are right angles. For the same reason the angles at the points B, C, D, are right angles. And because AEB is a right angle, also EBG is a right angle; therefore GH is parallel to AC. For the same reason AC is parallel to FK; wherefore also GH is parallel to FK. In like manner we show that each of the lines GF, HK, is parallel to BED. Therefore GK, GC, AK, FB, BK, are parallelograms; whence GF is equal to HK,d also GH to FK. And because AC is equal to BD, but also AC is equal to each of the lines GH, fk, and ED to each of the lines GF, HK, therefore GH, FK are equal to GF, HK, each to each. Whence the quadrilateral figure FGHK is equilateral. It is also rectangular. For because GBEA is a parallelogram, and AEB is a right angle, therefore AGB is a right angle. In like manner, we show that the angles at H, K, F, are right angles; therefore the quadrilateral figure FGHK is rectangular; and it has been shown to be equilateral; therefore it is a square. And it is circumscribed about Therefore about a given circle, the circle ABCD. &c. Q. E. F.* PROPOSITION VIII. PROBLEM. To inscribe a circle in a given square. Let ABCD be a given square; it is required to inscribe a circle in the square ABCD. F A B E H D K a A b 31. 1. e 34. 1. Bisect each of the lines AB, AD, in the points E, F," 10. 1. and through E draw EH parallel to either of them AB, CD; and through F draw FK parallel to either of them AD, BC; b therefore each of them KB, AH, HD, AG, GC, BG, GD, is a parallelogram, and their opposite sides are equal. And because AD is equal to AB, and AE is half of AD, also AF half of AB, therefore AE is equal to AF; and their opposite sides are equal, whence FG is equal to GE. In like manner we show that each of them FG, GE, is equal to each of them GH, GK; therefore the four GE, GF, GH, GK, are equal to one another. Whence with centre G, and distance any one of them GE, GF, GH, GK, a circle described will pass through the remaining points, and touch the right lines AB, BC, CD, DA, wherefore the angles at E, F, H, K, are right angles. For if the circle cut the lines AB, BC, CD, DA, the right line drawn from the extremity at right angles to the diameter of the circle, that right line will fall within the circle, which has been shown to be absurd.d Therefore with the centre G and dis- d 16. 3. tance any one of them GE, GF, GH, GK, a circle described will not cut the right lines AB, BC, CD, DA. Therefore it will touch them, and will be inscribed in the square ABCD. Therefore in a given square, &c. Q. E. F. Deduction. To inscribe a circle in a given rhombus. * If a regular polygon of any number of sides be inscribed in a circle, and it is required to circumscribe about that circle a regular polygon of the same number of sides, and reciprocally, the circumscribed polygon being given to construct the inscribed polygon. See Lacroix's Elémens de Géométrie, page 96. . 8. 2. b 6.1. a 11.2. PROPOSITION IX. PROBLEM. To circumscribe a circle about a given square. Let ABCD be a given square; it is required to circumscribe a circle about the square ABCD. For join AC, BD, which will cut one another in the point E. And because DA is equal to AB, and AC common, the two DA, AC, are equal to the two BA, AC, and the base DC will be equal to the base BC, and the angle DAC equal to the angle BAC; therefore the angle DAB is bisected by the right line AC. In like manner we E demonstrate, that each of the angles Deduction. To describe a circle about a given parallelogram. PROPOSITION X. PROBLEM. To make an isosceles triangle having each of the angles at the base double of the remaining angle. Let AB be any given right line, and cut it in the point c, so that the rectangle contained under AB, BC, may be equal to the square of CA, and with cenb 3 post. tre A, and distance AB, describe the circ 1. 4. cle BDE, also in the circle BDE apply the right line BD equal to the right line b AC, not greater than the diameter of the B E D |