Let a b, Fig. 32., be the circle drawn in perspective. It is required to divide it into a given number of equal parts; in this case, 20.

Let K a l be the semicircle used in the construction. Divide the semicircle K a l into half the number of parts required; in this case, 10.

Produce the line e g laterally, as far as may be necessary.

From o, the centre of the semicircle k a l, draw radii through the points of division of the semicircle, p, q, r, c, and produce them to cut the line E G in p, Q, R, c.

From the points p Q R draw the lines p p', Q q', R r', c, through the centre of the circle A B, each cutting the circle in two points of its circumference.

Then these points divide the perspective circle as required.

If from each of the points p, q, r, a vertical were raised to the line E G, as in Fig. 31., and from the point where it cut e g a line were drawn to the vanishing-point, as Q q' in Fig. 31., this line would also determine two of the points of division.

If it is required to divide a circle into any number of given unequal 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 x Y, 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 A B c, its representative point, a, must be on the nearer side of the circle a b c; and if the point B is on the farther side of the circle A b c, b

Fig. 32.

Fig. 33.

horizontal c P, find the correspondent p in the side of the perspective square, and draw p c parallel to x y, cutting the perspective circle in c.

It is obvious that if the points p', Q', r', c, 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

Let a b, Fig. 34., be the sight-magnitude of the side of the smaller square, and A c that of the side of the larger square.

Fig. 34.

Draw the larger square. Let d E F G be the square so drawn.

Join E G and D F.

On either d e or D g set off, in perspective ratio, D H equal to one half of B c. Through H draw H K to the vanishing-point of D E, cutting D F in I and E G in K. Through I and K draw I M, K L, to vanishing-point of D G, cutting d F in l and E G in M. Join L M.

Then I k l m is the smaller square, inscribed as required.1

Fig. 36.

If, instead of one square within another, it be required to draw one circle within another, the dimensions of both being given, enclose each circle in a square. Draw the squares first, and then the circles within, as in Fig. 36.

1 If either of the sides of the greater square is parallel to the plane of the picture, as D G in Fig. 35., d g of course must be equal to A c, and D H equal to BC/2 , and the construction is as in.

Fig. 35.