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... So on the line ZY, starting from point Z and going upwards, a sixth of ZY, and thus you will find point E. Point F will be on the drawing, in the same manner, by seeking with the pencil holder at what height on AB point H will be level with point F. Here point H is from point B, descending, at a large quarter of the total line AB. You will take on the drawing, from point Y and descending, the large quarter of the total line ZY, thus finding point F, you have completed the line EF, forming the junction of the two walls, or the corner of the room opposite the observer. The height EF thus found appears to be just half the heights AB and CD. You can verify whether this result is correct by comparing the entire line EF with AB and CD. By holding the pencil holder near your eye so that the upper end is level with the point represented by F and the level index with the point represented by E, you then transfer this height to AB or CD, keeping your hand well parallel to yourself, and you recognize that EF is indeed only half of AB and CD. The three lines AB, CD, and EF being thus found, you join the ends by lines represented here by BF, DF, AE, CE, yielding the large lines forming the entire interior to be drawn.

To draw the interior details, you must find the points of convergence. The observer being placed opposite the angle formed at EF, his viewpoint will be on this line, at a point V, at eye level, and noting that no vanishing line of the drawing converges at this point. It was seen when starting (page 25) that the lines that do not make a right angle with the ground line or the observer, converge to Accidental Points. Here, the vanishing lines of the interior not making a right angle with the ground line or observer, must converge to accidental points, which can be found with much ease. It suffices to extend on the drawing the lines represented here by BF, AE, until they meet at a point M. This point M will indeed be an accidental point where all the vanishing lines that will be found in this side of the interior and which are parallel to the lines represented by BF and AE will converge. To find the accidental point where all the vanishing lines on the other side of the interior will converge, extend on the drawing the lines represented by DF and CE. They will meet at a point L, which is the requested point of convergence.

The preceding demonstration relates to that of the box of plate 3, figure 12: the room in question is like the interior of the box. In the other application, as it is the exterior of the box that is presented, the middle angle being closer to the eye, appears larger than the others. Here, this angle, being farther away, appears smaller.

Once the entire interior is found, deal with the details. If there is, for example, a door in this right side of the room, to draw this door in perspective and in relation to the wall, proceed as follows. Hold the pencil holder horizontally so that the end of the pencil is level with a point on line PR marking the first side of the door from that point, and placing the index at level with the point where the pencil holder then intersects line CD, you have on the pencil holder the distance that is from the beginning of the wall on the right side to the start of the door on that same side. Compare this distance to the total width of the wall. This distance is here almost half of the width of the wall CDFE. Divide, on the drawing, this wall CDFE into two nearly equal parts, and, through the point marking this division, draw a vertical line which is represented here by PR. With the pencil holder, evaluate the distance between the top of the wall and the top of the door, represented here by P. The distance P is contained seven times in the height JR. Therefore, take, on the drawing, and on the vertical JR, from J to R, one-seventh of JR, and thus find point P. Draw from this point a line to point L, which is the point of convergence for this side. Then, still with the pencil holder, compare the width of the door with the width of the wall, from PR to DC or to FE, finding here that the width PS or RT, taken at the top, in the middle, or at the bottom, equals half the distance from PR to DC or to FE. Therefore, divide on the