D C are together equal to the sides AD and BC, which W. T. B. D. Inversely If two opposite sides of a quadrangle be not together equal to the two others, the quadrangle is not circumscriptible to a circle. THEOREM 91. Conversely, if two opposite sides of a convex quadrangle be together equal to the others, the quadrangle is circumscriptible to the circle. Let A B C D be a convex quadrangle having the opposite sides A B and D C equal together to the two others A D and B C. Let BE and CF be the bisectrices of the angles B and C. Because the angles A, B, C and D are together equal to four right angles [47], the angles B and C are together less than four right angles; hence their halves EBC and FCB are together less than two right angles [vii]: therefore the lines B E and CF are oblique to each other [II. 20 ii] and, infinitely produced, will meet [I. 18 i]. Let O be the intersection of the lines B E and C F; let O G, O H and OI be perpendicular to the sides A B, BC and CD respectively. Because the point O is on the bisectrix of the angle B, the line O G is equal to O H [II. 10]; because the same point is on the bisectrix of the angle C, the line O I is equal to O H; therefore O G is equal to O I [i]. Let a circumference be drawn from the centre O with OG as a radius; then it will pass through the points H and I [I. Def. 23]. Because the sides BA, BC and CD are perpendicular at the extremities of the radii O G, OH and OI respectively, they are tangent to the circle O [I. 54]; hence B G is equal to BH [I. 57], and CI is equal to CH; therefore B G and CI are together equal to BH and HC, or BC [iv]: but A B and DC are together equal to B C and AD [Hyp.]; therefore A G and DI are together equal to A D [iv]. From the centre A let a circumference be drawn through the point G, and from the centre D let a circumference be drawn through the point I; then, because their centre line is equal to the sum of their radii [Dem.], these two circumferences will be externally tangent to each other in a point K [I. 62]. Because O G and O I are perpendicular at the extremities of the radii AG and DI respectively, they are tangent to the circumferences A and D respectively: because O G and OI are equal to each other [Dem.], the circumference O is tangent to AD in the point K [I. 67 ii]: therefore the quadrangle A B C D is circumscribed to the circum. ference O, which W. T. B. D. COROLLARY. Every lozenge is circumscriptible to a circle. THEOREM 92. Every regular polygon is inscriptible in a circle. Because the vertices of any regular polygon are equally distant from the centre of regularity, the circumference drawn from the latter point as a centre, through one of the vertices, will pass through all the others; therefore every regular polygon is inscriptible in a circle, W. W. T. B. D. Inversely Any polygon which is not inscriptible in a circle is not regular. COROLLARY I. The centre of the circle circumscribed to a regular polygon, is the centre of regularity of the inscribed polygon. COROLLARY II. The radius of the circle circumscribed to a regular polygon, is equal to the radius of the inscribed polygon. COROLLARY III. The points of division of a circumference divided into more than two equal parts, are the vertices of a regular polygon. COROLLARY IV. If a regular polygon be inscribed in a circle, the extremities and the middle points of the arcs subtended by the sides, are the vertices of a regular polygon of double the number of sides. Scholium. The sides of a regular polygon are equally distant from the centre (55 ii). THEOREM 93. Every regular polygon is circumscriptible to a circle. Because the apothems of a regular polygon are equal to one another [55 ii], the circumference drawn from the centre of regularity with one of them as a radius, will be tangent to all the sides of the polygon [I. 54]; therefore every regular polygon is circumscriptible to a circle, W. W. T. B. D. Inversely Any polygon which is not circumscriptible to a circle is not regular. COROLLARY I. The centre of the circle inscribed in a regular polygon is the centre of regularity of the circumscribed polygon. COROLLARY II. The radius of the circle inscribed in a regular poly. gon, is equal to the apothem of the circumscribed polygon. COROLLARY III. Two adjacent sides of a regular polygon, are symmetric to the radius terminating at their common extremity (I. 57 iv). Scholium. The circle circumscribed to a polygon and the circle inscribed therein, are concentric. THEOREM 94. If a circumference be divided into more than two equal parts, the intersections of the straight lines tangent thereto through the points of division, are the vertices of a regular meet in the points E, I, K, etc., being the vertices of a polygon P. Let A B, BC and CD be joined; then the lines A B, B C and CD are equal to one another [I. 42]. Because the circumscribed angles AHB and BIC have equal convex arcs, they are equal to each other [II. 30 iii]: likewise the angle CKD and the remaining angles thus formed, are equal to the angle AH B; therefore the polygon P is equiangular. Because AH is equal to H B [I. 57], BI equal to I C, and C K equal to K D, the triangles AH B, BIC and CKD are isosceles. Because the isosceles triangles AHB, BIC and CKD have equal bases and equal angles at the apex [Dem.], they are equal to each other [16 iv]; thence the lines AH, HB, BI, IC, C K and K D are equal to each other: therefore H I is equal to IK [iv]. The same construction and reasoning would prove that the remaining sides of the polygon P are each equal to HI; then the polygon P is equilateral: but it is also equiangular [Dem.]; therefore the polygon P is regular [58], W. W. T. B. D. COROLLARY. The tangents through the extremities of two perpendicular diameters, form a square. E C PROBLEMS. PROBLEM 1. Given two sides of a triangle and the angle which the form, to construct the triangle. Let A B and C D be the given sides, and let E be the given angle D F A At the point A draw A F making with A B an angle B A F equal to the given angle E [II. Prob. 1]; take A H equal to CD, and join B H: the triangle A B H is the required triangle. For the side A H is equal to CD [Const.], and the angle B A H, formed by the sides A B and A H, is equal to the given angle E [Const.], which was to be done. PROBLEM 2. Given one side of a triangle and the two adjacent angles, to construct the triangle. PROBLEM 3. (Eucl. I. 1.) Upon a given straight line to construct an equilateral triangle. Let A B be the given straight line. From the centres A and B, with a radius equal to A B, describe |