EGB, will be equal, since they have the hypotenuse and a side of the one respectively equal to the hypotenuse and a side of the other (B. I., T. XXVI.), and BF = BG. In a similar manner we may show that AF-AK; CH=CG; DH = DK. Consequently by adding, we have AB+CD = BC+AD. · Cor. When the sums of the opposite sides of a quadrilateral are equal, it is capable of circumscribing a circle. For, if we describe a circumference tangent to AD, AB, and BC, which can always be done (T. XII., S. I. or II.), it will also be tangent to DC. For, if the side DC is not tangent to the circumference, it must be either a secant or lie wholly without the circumference. We will first suppose it to be a secant. If DE A B D CE C is drawn tangent to this circumference, we shall have AD+BE = AB+DE, but by hypothesis we have AD+BC= AB+ DC; hence, by subtraction, we obtain EC=DE - DC; that is, one side of a triangle is equal to the difference of the other two sides, which is impossible (B. I., T. VII.). Hence the side DC cannot be secant. In a similar way we can show that it cannot be wholly without the circumference. It must therefore be tangent. Scholium. The rhombus and the square are the only quadrilaterals capable of circumscribing a circle. THEOREM XV. All regular polygons are capable of being inscribed in a circle, and of circumscribing a circle. Let ABCDEF be any regular polygon; and through three consecutive angles A, B, and C, describe the circumference of a circle (T. XII.). This circumference will also pass through all the other angles of the polygon. For, drawing from the centre G, the lines GA, GB, GC, and GD, we shall form three triangles, GAB, GBC, GCD. In the two triangles GAB and GBC, we have GB common, AB = BC and GA= GC; hence the triangles are equal (B. I., T. XXV.), and being isosceles since GA = GB=GC, we have the angles GAB=GBA=GBC= GCB. That is, each of these angles is one half the angle of the polygon; hence the angle GCBGCD. Now, comparing the two triangles GBC and GCD, we have the side GC common, BC = CD, and the angle GCB= GCD; these triangles are therefore equal (B. I., T. XX.); and we have GD = GB=GC=GA; hence, the circumference which passes through A, B, and C, will also pass through D. In a similar manner we can show that the circumference which passes through B, C, and D, will also pass through E, and so on, for all the angles of the polygon, which demonstrates the first part of the Theorem. As to the second part, we observe, that the sides AB, BC, CD, etc., are equal chords, in reference to the circumscribed circle, and therefore they are equally distant from the centre G (T. VII., C. I.). If then, with G as a centre, and with the perpendicular GH as a radius, we describe a circumference, it will be tangent to all the sides of the polygon. This circle will be inscribed in the polygon, and the polygon will circumscribe the circle. Scholium. The point F, the common centre of the inscribed and circumscribed circles, is called the centre of the regular polygon. The radius of the circumscribing circle is called the radius of the polygon, and the radius of the inscribed circle is called the apothem. The angles AGB, BGC, CGD, etc., are called angles at the centre. The lines GA, GB, GC, etc., which are the radii of the polygon, bisect the angles of the polygon. The apothems GH, GK, GL, etc., which are the radii of the inscribed circle, bisect the sides of the polygon perpendicularly, and also bisect the angles at the centre, since they would, if produced, bisect the arcs which measure the angles at the centre. THEOREM XVI. If through the corners of a regular polygon, already inscribed in a circle, we draw tangents, we shall thus circumscribe a regular polygon of the same number of sides. = = A" E' A' B' B The tangents thus drawn will form with the sides of the inscribed polygon, AB, BC, CD, etc., a series of triangles, AA'B, BB'C, CC'D, etc., all isosceles and equal, since AB BC = CD = etc., and the angles A'AB, A'BA, B'BC, B'CB, C’CD, C'DC, etc., are all equal, having for their measure half of the equal arcs AB, BC, CD, etc. (T. X., C. II.). Hence, all the angles, A', B', C', etc., of the circumscribed polygon are equal. And since AA'=A'B=BB'B'C = CC' etc., we have A'B'B'C' C'D' etc. That is, this circumscribed polygon has all its angles equal, and all its sides equal; it is therefore regular (D. VI.). = = D' = Scholium. Since the radius OA' of the circumscribed polygon bisects the angle E'A'B' (T. XV., S.), we have, in the two rightangled triangles OAA' and OBA', the acute angle OA'A of the one, equal to the acute angle OA'B of the other; consequently, the remaining acute angles are equal; that is, the angle A'OA=A'OB. In the same way we can show that the angle B'OB=B'OC. But the angle A'OB = B'OB, since the apothem FB bisects the angle A'FB' at the centre (T. XV., S.). Hence, all the angles AOA', A’OB, BOB', B'OC, COC', etc., are equal; and their measuring arcs AI, IB, BK, KC, CL, etc., are all equal. If then we suppose the circumscribed polygon to revolve in its own plane about its centre O, as a pivot, so that A may reach the point I, B will then coincide with K, C with L, etc. The circumscribed polygon will by this means, without having its relative parts in the least changed, assume a new position, having its sides parallel with the sides of the inscribed polygon. OF SECANT AND TANGENT CIRCLES. THEOREM XVII. If two circumferences of circles have two points in common, the line joining their centres will bisect their common chord perpendicularly. For, the line bisecting a chord at right angles passes through the centre (T. IV., S.); and as this chord is common to both circles, this bisecting line must pass through both centres. THEOREM XVIII. When two circumferences have only one point in common, this point will be situated on the line joining their centres. A E B D If A and B are the centres of two circles having only one point in common, that point will be situated on the line AB joining their centres. For, if not, suppose it to be situated without this line, as at C. Draw CE perpendicular to AB, and prolong it until ED CE; then we evidently have AC AD, also BC=BD, and since C was a common point of the circumferences, D will also be a common point. These circumferences will then have two points in common, which is contrary to the hypothesis. Hence, the point in common to the two circumferences must be in the line joining their centres. = = Scholium. Suppose we consider two circles whose centres are at A and B, the one at B being the smaller. Draw CF passing through their centres, and at the extremities of their diameters, CD and EF, draw the perpendiculars GG', HH', KK', LL'. If we suppose the circle, whose centre is at A, fixed, and the other to move towards it, so that the centre B may move in the line AB, we may notice five distinct positions of these circles. |