PROP. III. PROBLEM. To circumscribe a triangle about a given circle, that shall be equiangular to a given triangle. M D K Let ABC be the given circle, and der the given triangle; it is required to circumscribe a triangle about the circle ABC that shall be equiangular to the triangle DEF. Produce the line EF to G and H; and, at the centre 1, make the angles AIB, Bic equal to the angles DEG, DFH (I. 20.); and draw the lines MK, KL, LM, to touch the circle in the points A, B, C (III. 10.); and join AB. Then, since the angles IAK, KBI, are, each of them, a right angle (III. 12.), the angles BAK, KBA, taken together, will be less than two right angles. But when a right line intersects two other right lines, and makes the two interior angles, on the fame side, together less than two right angles, those lines will, if produced, meet each other (I. 25. Cor.) The line MK, therefore, meets the line Kl; and, if A, C, C, B be joined, the same may be proved of the lines KL, LM and MK; consequently the figure Kem is a triangle. And, because the four angles of the quadrilateral AIBK. are equal to four right angles (Cor. I. 28.), and the angles 13 IAK, ! JAK, KBI are each a right angle, the remaining angles AIB, BKA will be equal to two right angles. But the angles DEG, DEF are also equal to two right angles (I. 13.); therefore, since the angle deg is equal to the angle AIB (by Conft.), the remaining angle BKA will be equal to the remaining angle der. And, in the fame manner, it may be proved, that the angle CLB is equal to the angle die. The angle MKL, therefore, being equal to the angle DeF, and the angle Mlk to the angle dfe, the remaining angle KML will also be equal to the remaining angle EDF; and consequently the triangle klm is equiangular to the triangle eFD. Q. E. D. PROP. IV. PROBLEM. In a given triangle to inscribe a circle. G B Let ABC be the given triangle; it is required to inscribe zcircle in it. Bisect the angles CAB, ABC, with the right lines AD, DB (I. 9.) Then, since the angles DAB, DBA are less than two right angles (I. 28.), the lines AD, DB, will, if produced, meet each other (I. 25. Cor.) And, if from the point of intersection D, there be drawn the perpendiculars DF, DG and De, they will be the radii of the circle required. 1 For, since the angle EaD is equal to the angle DAF (by Conft.), and the angle Aed to the angle DFA, (being each of them right angles), the remaining angle eda will also be equal to the remaining angle ADF (I. 28. Cor.) The triangles ADE, DAF, therefore, being equiangular, and having the side ad common to both, the side de will also be equal to the side DF (I. 21.) And, in the same manner, it may be proved, that the side DG is equal to the side DF. The right lines DE, DG and DF are, therefore, all equal to each other; and the angles at the points F, E and G are right angles, by construction. If, therefore, a circle be described from the centre D, with either of the distances DE, DG or DF, it will touch the fides in the points E, G, F (III. 10.) and be inscribed in the triangle ABC, as was to be done. PRO P. V. PROBLEM, To circumscribe a circle about a given triangle, Let ABC be the given triangle; it is required to circumscribe a circle about it. Bifect the fides AC, CB with the perpendiculars DE, EF (I. ļo and 11.); and join DF, Then, since the angles EDF, DFE are less than two right angles (by Conft.), the lines DE, EF will meet each other (1. 25. Cor.) Let E, therefore, be their point of intersection, and draw the lines EA, EC and EF. Then, because AD is equal to DC (by Conf.), de common, and the angle Ade equal to the angle EDC (being each of them right angles), the base AE will also be equal to the base ec (I. 4.) And, in the same manner, it may be proved, that ec is equal to EB; consequently EA, EC and EB are all equal to each other. If, therefore, a circle be described from the point E, at either of the distances EA, EC or EB, it will pass through the remaining points, and circumscribe the triangle ABC, as was to be done. PRO P. VI. PROBLEM. To inscribe a square in a given circle. B Let ABCD be the given circle; it is required to inscribe a square in it. Through £, the centre of the circle, draw any two diameters AC, BD at right angles to each other (I. 11, 12.), and join AB, BC, CD and DA; then will bcda be the square required. For fince the two sides be, EA, are equal to the two fides ED, EA, and the angle bea to the angle AED, (being each of them right angles), the base BA will be equal to the base AD (I. 4.) And, in the same manner, it may be proved, that the fides BC, cd are each equal to the sides BA, AD; whence the figure BCDA is equilateral. It is also rectangular : for fince bda is a semi-circle, the angle bad is a right angle (III. 12.) And, for the fame reason, the angles ABC, BCD and CDA are each of them right angles. The figure BCDA, therefore, being equilateral, and having all its angles right angles is a square, and it is infcribed in the circle ABCD, as was to be done. PRO P. VII. PROBLEM. To circumscribe a square about a given circle. F B E Let ABCD be the given circle; it is required to circumscribe a square about it. Draw any two diameters AC, BD at right angles to each other (I. 11, 12.); and through the points A, B, C, D, draw the tangents KF, FG, GH, HK (III. 10.); and join AB. Then, |