If it is required to divide a circle into any number of given _un_equal parts (as in the points _A_, _B_, and _C_, Fig. 33.), the shortest way is thus to raise vertical lines from _A_ and _B_ to the side of the perspective square _XY_, and then draw to the vanishing-point, cutting the perspective circle in _a_ and _b_, the points required. Only notice that if any point, as _A_, is on the nearer side of the circle _ABC_, its representative point, _a_, must be on the nearer side of the circle _abc_; and if the point _B_ is on the farther side of the circle _ABC_, _b_ must be on the farther side of _abc_. If any point, as _C_, is so much in the lateral arc of the circle as not to be easily determinable by the vertical line, draw the horizontal _CP_, find the correspondent _p_ in the side of the perspective square, and draw _pc_ parallel to _XY_, cutting the perspective circle in _c_. [Illustration: Fig. 33.] COROLLARY. It is obvious that if the points _P′_, _Q′_, _R_, etc., by which the circle is divided in Fig. 32., be joined by right lines, the resulting figure will be a regular equilateral figure of twenty sides inscribed in the circle. And if the circle be divided into given unequal parts, and the points of division joined by right lines, the resulting figure will be an irregular polygon inscribed in the circle with sides of given length. Thus any polygon, regular or irregular, inscribed in a circle, may be inscribed in position in a perspective circle. PROBLEM XIII. TO DRAW A SQUARE, GIVEN IN MAGNITUDE, WITHIN A LARGER SQUARE GIVEN IN POSITION AND MAGNITUDE; THE SIDES OF THE TWO SQUARES BEING PARALLEL. [Illustration: Fig. 34.] Let _AB_, Fig. 34., be the sight-magnitude of the side of the smaller square, and _AC_ that of the side of the larger square. Draw the larger square. Let _DEFG_ be the square so drawn. Join _EG_ and _DF_. On either _DE_ or _DG_ set off, in perspective ratio, _DH_ equal to one half of _BC_. Through _H_ draw _HK_ to the vanishing-point of _DE_, cutting _DF_ in _I_ and _EG_ in _K_. Through _I_ and _K_ draw _IM_, _KL_, to vanishing-point of _DG_, cutting _DF_ in _L_ and _EG_ in _M_. Join _LM_. Then _IKLM_ is the smaller square, inscribed as required.[23] COROLLARY. [Illustration: Fig. 36.] If, instead of one square within another, it be required to draw one circle within another, the dimensions of both being given, inclose each circle in a square. Draw the squares first, and then the circles within, as in Fig. 36. [23] [Illustration: Fig. 35.] If either of the sides of the greater square is parallel to the plane of the picture, as _DG_ in Fig. 35., _DG_ of course must be equal to _AC_, and _DH_ equal to _BC_/2, and the construction is as in Fig. 35. PROBLEM XIV. TO DRAW A TRUNCATED CIRCULAR CONE, GIVEN IN POSITION AND MAGNITUDE, THE TRUNCATIONS BEING IN HORIZONTAL PLANES, AND THE AXIS OF THE CONE VERTICAL. Let _ABCD_, Fig. 37., be the portion of the cone required.