Problems. Let C be the centre of the circle, and A the given point without the circle. Join A and the centre C, and on AC, as a diameter, describe a circumference. Through the points B and D where the two circumferences intersect each other, draw the lines AB and AD: these lines will be tangent to the circle whose centre is C. For, since the angles ABC and B ADC are each inscribed in a semicircle, they will be right angles (Bk. II. Th. x). Again, since the lines AB, AD, are each perpendicular to a radius at its extremity, they will be tangent to the circle (Bk. II. Th. v). PROBLEM XIX To inscribe a circle in a given triangle. Let ABC be the given triangle. Bisect the angles A and B by the lines AO and BO, meeting at the point 0. From O, let fall the perpendiculars OD, OE, OF, on the three sides of A D B E the triangle these perpendiculars will be equal to each other. For, in the two right angled triangles DAO and FAO, we have the right angle D equal the right angle F, the angle FAO equal to DAO, and consequently, the third angles AOD and AOF are equal (Bk. I. Th. xvii. Cor 1). But the two common side AO, hence, they are equal triangles have a (Bk. I. Th. v), and consequently, OD is e ual to OF Problems. In a similar manner, it may be proved that OE and OD are equal: hence, the three perpendiculars, OD, OF, and OE, are all equal. Now, if with O as a centre, and OF as a radius, we describe D B F the circumference of a circle, it will pass through the points D and E. and since the sides of the triangle are perpendicular to the radii OF, OD, OE, they will be tangent to the circumference (Bk. II. Th. v). Hence, the circle will be inscribed in the triangle. PROBLEM XX. To inscribe an equilateral triangle in a circle. Through the centre C draw any diameter, as ACB. From B as a centre, with a radius equal to BC, describe the arc DCE. Then, draw AD, AE, and DE, and DAE will be the required triangle. D For, since the chords BD, BE, are each equal to the radius CB, the arcs BD, BE, are each equal to sixty degrees (Rk. II. Th. XIX), and the arc DBE to one hundred and twenty degrees; hence, the angle DAE is equal to sixty degrees (Bk. II. Th. viii). Again, since the arc BD is equal to sixty degrees, and the arc BDA equal to one hundred and eighty degrees, it follows that DA will be equal to one hundred and twenty degrees: hence, the angle DEA is equal to sixty degrees, and consequently, the third angle ADE, is equal to sixty degrees. Problems. Therefore, the triangle ADE is equilateral (Bk. I. Th. vi Cor. 2). PROBLEM XXI. To inscribe a regular hexagon in a circle. Draw any radius, as AC. Then apply the radius AC around the circum ference, and it will give the chords AD, H DE, EF, FG, GH, and HA, which will be the sides of the regular hexagon. For, F the side of a hexagon is equal to the radius (Bk. II. Th. xix). To inscribe a square in a given circle. Let ABCD be the given circle. Draw the two diameters AC, BD, at righ angles to cach other, and through the points A, B, C and D draw the lines AB, BC, CD, and DA: then will ABCD be the required square. For, the four right angled triangles, AOB, BOC, COD, and DOA are B D equal, since the sides AO, OB, OC, and OD are equal, being radii of the circle; and the angles at O are equal in each, being right angles: hence, the sides AB, BC, CD, and DA are equal (Bk. I. Th. iv). But each of the angles ABC, BCD, CDA, DAB, is a right angle, being an angle in a semicircle (Bk. II. Th x): hence, the figure ABCD is a square (Bk. I. De'. 48) Problems. Sch. If we bisect the arcs AB, BC, CD, DA, and join the points, we shall have a regular octagon inscribed in the circle. If we again B bisect the arcs, and join the points of bisection, we shall have a regular polygon of sixteen sides. D PROBLEM XXIII. To describe a square about a given circle. Draw the diameters AB, DE, at right angles to each other. Through the extremities A and B draw FAG and HBI parallel to DE, and through E and D, draw FEH and GDI parallel to AB: then will FGIH be the required square. A G E H B For, since ACDG is a parallelogram, the opposite sides are equal (Bk. I. Th. xxiii): and since the angle at C is a right angle. all the other angles are right angles (Bk. I. Th. xxiii. Cor. 1): and as the same may be proved of each of the figures CI, CH and CF, it follows that all the angles, F, G, I, ard H, are right angles, and that the sides GI, IH, HF, and FG, are equal, each being equal to the diameter of the circle. Hence the figure GIHF is a square (Bk. I. Def. 48). GEOMETRY. BOOK III. OF RATIOS AND PROPORTIONS. DEFINITIONS. 1. Ratio is the quotient arising from dividing one quantity by another quantity of the same kind. Thus, if the numbers 3 and 6 have the same unit, the ratio of 3 to 6 will be expressed by 6 And in general, if A and B represent quantities of the same kind, the ratio of A to B will be expressed by B 2. If there be four numbers, 2, 4, 8, 16, laving such values that the second divided by the first is equal to the fourth divided by the third, the numbers are said to be in proportion. And in general, if there be four quantities A, B, C, and D, having such values that B D then, A is said to have the same ratio to B, that C has to D, or, the ratio of A to B is equal to the ratio of C to D When |