It is interesting to notice that the solution tacitly assumes that a certain arc is going to cut the given line in two points, and only two. Strictly speaking, why may it not cut it in only one point, or even in three points? We really assume that if a straight line is drawn througha point within a circle, this line must get out of the circle on each of two sides of the given point, and in getting out it must cut the circle twice. Proclus noticed this assumption and endeavored to prove it. It is better, however, not to raise the question with beginners, since it seems to them like hair-splitting to no purpose.
The problem is of much value in surveying, and teachers would do well to ask a class to let fall a perpendicular to the edge of a sidewalk from a point 20 feet from the walk, using an ordinary 66-foot or 50-foot tape. Practically, the best plan is to swing 30 feet of the tape about the point and mark the two points of intersection with the edge of the walk. Then measure the distance between the points and take half of this distance, thus fixing the foot of the perpendicular.
Problem. At a given point in a line, to erect a perpendicular to that line.
This might be postponed until after the problem to bisect an angle, since it merely requires the bisection of a straight angle; but considering the immaturity of the average pupil, it is better given independently. The usual case considers the point not at the extremity of the line, and the solution is essentially that of Euclid. In practice, however, as for example in surveying, the point may be at the extremity, and it may not be convenient to produce the line.
Surveyors sometimes measure PB = 3 ft., and then take 9 ft. of tape, the ends being held at B and P, and the tape being stretched to A, so that PA = 4 ft. and AB = 5 ft. Then P is a right angle by the Pythagorean Theorem. This theorem not having yet been proved, it cannot be used at this time.
A solution for the problem of erecting a perpendicular from the extremity of a line that cannot be produced, depending, however, on the problem of bisecting an angle, and therefore to be given after that problem, is attributed by Al-Nairīzī (tenth century A.D.) to Heron of Alexandria. It is also given by Proclus.
Required to draw from P a perpendicular to AP. Take X anywhere on the line and erect XY ⊥ to AP in the usual manner. Bisect ∠PXY by the line XM. On XY take XN = XP, and draw NM ⊥ to XY. Then draw PM. The proof is evident.