We now have to find certain points by which to measure those vanishing or retreating lines which are no longer at right angles to the picture plane, as in parallel perspective, and have to be measured in a different way, and here geometry comes to our assistance.
Fig. 116.
Note that the perspective square P equals the geometrical square K, so that side AB of the one equals side ab of the other. With centre A and radius AB describe arc Bm· till it cuts the base line at m·. Now AB = Am·, and if we join bm· then triangle BAm· is an isosceles triangle. So likewise if we join m·b in the perspective figure will m·Ab be the same isosceles triangle in perspective. Continue line m·b till it cuts the horizon in m, which point will be the measuring point for the vanishing line AbV. For if in an isosceles triangle we draw lines across it, parallel to its base from one side to the other, we divide both sides in exactly the same quantities and proportions, so that if we measure on the base line of the picture the spaces we require, such as 1, 2, 3, on the length Am·, and then from these divisions draw lines to
the measuring point, these lines will intersect the vanishing line AbV in the lengths and proportions required. To find a measuring point for the lines that go to the other vanishing point, we proceed in the same way. Of course great accuracy is necessary.
Note that the dotted lines 1,1, 2,2, &c., are parallel in the perspective, as in the geometrical figure. In the former the lines are drawn to the same point m on the horizon.
[ LVIII]
How to Divide any Given Straight Line into Equal or Proportionate Parts
Let AB (Fig. 117) be the given straight line that we wish to divide into five equal parts. Draw AC at any convenient angle, and measure off five equal parts with the compasses thereon, as 1, 2, 3, 4, 5. From 5C draw line to 5B. Now from each division on AC draw lines 4, 4, 3, 3, &c., parallel to 5,5. Then AB will be divided into the required number of equal parts.