Book VI, Proposition 33. Such a procedure is not adapted to the needs of

students to-day. Euclid gave in Book III, however, the proposition (No. 20) that a central angle is twice an inscribed angle standing on the same arc. Since Euclid never considered an angle greater than 180°, his inscribed angle was necessarily less than a right angle. The first one who is known to have given the general case, taking the central angle as being also greater than 180°, was Heron of Alexandria, probably of the first century A.D.[68] In this he was followed by various later commentators, including Tartaglia and Clavius in the sixteenth century.

One of the many interesting exercises that may be derived from this theorem is seen in the case of the "horizontal danger angle" observed by ships.

If some dangerous rocks lie off the shore, and _L_ and _L'_ are two lighthouses, the angle _A_ is determined by observation, so that _A_ will lie on a circle inclosing the dangerous area. Angle _A_ is called the "horizontal danger angle." Ships passing in sight of the two lighthouses _L_ and _L'_ must keep out far enough so that the angle _L'SL_ shall be less than angle _A_.

To this proposition there are several important corollaries, including the following:

1. _An angle inscribed in a semicircle is a right angle._ This corollary is mentioned by Aristotle and is attributed to Thales, being one of the few propositions with which his name is connected. It enables us to describe a circle by letting the arms of a carpenter's square slide along two nails driven in a board, a pencil being held at the vertex.

A more practical use for it is made by machinists to determine whether a casting is a true semicircle. Taking a carpenter's square as here shown, if the vertex touches the curve at every point as the square slides around, it is a true semicircle. By a similar method a circle may be described by sliding a draftsman's triangle so that two sides touch two tacks driven in a board.

Another interesting application of this corollary may be seen by taking an ordinary paper protractor _ACB_, and fastening a plumb line at _B_. If the protractor is so held that the plumb line cuts the semicircle at _C_, then _AC_ is level because it is perpendicular to the vertical line _BC_. Thus, if a class wishes to determine the horizontal line _AC_, while sighting up a hill in the direction _AB_, this is easily determined without a spirit level.

It follows from this corollary, as the pupil has already found, that the mid-point of the hypotenuse of a right triangle is equidistant from the three vertices. This is useful in outdoor measuring, forming the basis of one of the best methods of letting fall a perpendicular from an external point to a line.

Suppose _XY_ to be the edge of a sidewalk, and _P_ a point in the street from which we wish to lay a gas pipe perpendicular to the walk. From _P_ swing a cord or tape, say 60 feet long, until it meets _XY_ at _A_. Then take _M_, the mid-point of _PA_, and swing _MP_ about _M_, to meet _XY_ at _B_. Then _B_ is the foot of the perpendicular, since [L]_PBA_ can be inscribed in a semicircle.

2. _Angles inscribed in the same segment are equal._

By driving two nails in a board, at _A_ and _B_, and taking an angle _P_ made of rigid material (in particular, as already stated, a carpenter's square), a pencil placed at _P_ will generate an arc of a circle if the arms slide along _A_ and _B_. This is an interesting exercise for pupils.

THEOREM. _An angle formed by two chords intersecting within the circle is measured by half the sum of the intercepted arcs._

THEOREM. _An angle formed by a tangent and a chord drawn from the point of tangency is measured by half the intercepted arc._

THEOREM. _An angle formed by two secants, a secant and a tangent, or two tangents, drawn to a circle from an external point, is measured by half the difference of the intercepted arcs._

These three theorems are all special cases of the general proposition that the angle included between two lines that cut (or touch) a circle is measured by half the sum of the intercepted arcs. If the point passes from within the circle to the circle itself, one arc becomes zero and the angle becomes an inscribed angle. If the point passes outside the circle, the smaller arc becomes negative, having passed through zero. The point may even "go to infinity," as is said in higher mathematics, the lines then becoming parallel, and the angle becoming zero, being measured by half the sum of one arc and a negative arc of the same absolute value. This is one of the best illustrations of the Principle of Continuity to be found in geometry.

PROBLEM. _To let fall a perpendicular upon a given line from a given external point._

This is the first problem that a student meets in most American geometries. The reason for treating the problems by themselves instead of mingling them with the theorems has already been discussed.[69] The student now has a sufficient body of theorems, by which he can prove that his constructions are correct, and the advantage of treating these constructions together is greater than that of following Euclid's plan of introducing them whenever needed.

Proclus tells us that "this problem was first investigated by Oenopides,[70] who thought it useful for astronomy." Proclus speaks of such a line as a gnomon, a common name for the perpendicular on a sundial, which casts the shadow by which the time of day is known. He also speaks of two kinds of perpendiculars, the plane and solid, the former being a line perpendicular to a line, and the latter a line perpendicular to a plane.

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 through a 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-Nair[=i]z[=i] (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_ [perp] to _AP_ in the usual manner. Bisect [L]_PXY_ by the line _XM_. On _XY_ take _XN_ = _XP_, and draw _NM_ [perp] to _XY_. Then draw _PM_. The proof is evident.

These may at the proper time be given as interesting variants of the usual solution.

PROBLEM. _To bisect a given line._

Euclid said "finite straight line," but this wording is not commonly followed, because it will be inferred that the line is finite if it is to be bisected, and we use "line" alone to mean a straight line. Euclid's plan was to construct an equilateral triangle (by his Proposition 1 of Book I) on the line as a base, and then to bisect the vertical angle. Proclus tells us that Apollonius of Perga, who wrote the first great work on conic sections, used a plan which is substantially that which is commonly found in textbooks to-day,--constructing two isosceles triangles upon the line as a common base, and connecting their vertices.

PROBLEM. _To bisect a given angle._

It should be noticed that in the usual solution two arcs intersect, and the point thus determined is connected with the vertex. Now these two arcs intersect twice, and since one of the points of intersection may be the vertex itself, the other point of intersection must be taken. It is not, however, worth while to make much of this matter with pupils. Proclus calls attention to the possible suggestion that the point of intersection may be imagined to lie outside the angle, and he proceeds to show the absurdity; but here, again, the subject is not one of value to beginners. He also contributes to the history of the trisection of an angle. Any angle is easily trisected by means of certain higher curves, such as the conchoid of Nicomedes (_ca._ 180 B.C.), the quadratrix of Hippias of Elis (_ca._ 420 B.C.), or the spiral of Archimedes (_ca._ 250 B.C.). But since this problem, stated algebraically, requires the solution of a cubic equation, and this involves, geometrically, finding three points, we cannot solve the problem by means of straight lines and circles alone. In other words, the trisection of _any_ angle, by the use of the straightedge and compasses alone, is impossible. Special angles may however be trisected. Thus, to trisect an angle of 90° we need only to construct an angle of 60°, and this can be done by constructing an equilateral triangle. But while we cannot trisect the angle, we may easily approximate trisection. For since, in the infinite geometric series 1/2 + 1/8 + 1/32 + 1/128 + ..., _s_ = _a_ ÷ (1 - _r_), we have _s_ = 1/2 ÷ 3/4 = 2/3. In other words, if we add 1/2 of the angle, 1/8 of the angle, 1/32 of the angle, and so on, we approach as a limit 2/3 of the angle; but all of these fractions can be obtained by repeated bisections, and hence by bisections we may approximate the trisection.

The approximate bisection (or any other division) of an angle may of course be effected by the help of the protractor and a straightedge. The geometric method is, however, usually more accurate, and it is advantageous to have the pupils try both plans, say for bisecting an angle of about 49 1/2°.

Applications of this problem are numerous. It may be desired, for example, to set a lamp-post on a line bisecting the angle formed by two streets that come together a little unsymmetrically, as here shown, in which case the bisecting line can easily be run by the use of a measuring tape, or even of a stout cord.

A more interesting illustration is, however, the following:

Let the pupils set a stake, say about 5 feet high, at a point _N_ on the school grounds about 9 A.M., and carefully measure the length of the shadow, _NW_, placing a small wooden pin at _W_. Then about 3 P.M. let them watch until the shadow _NE_ is exactly the same length that it was when _W_ was fixed, and then place a small wooden pin at _E_. If the work has been very carefully done, and they take the tape and bisect the line _WE_, thus fixing the line _NS_, they will have a north and south line. If this is marked out for a short distance from _N_, then when the shadow falls on _NS_, it will be noon by sun time (not standard time) at the school.

PROBLEM. _From a given point in a given line, to draw a line making an angle equal to a given angle._

Proclus says that Eudemus attributed to Oenopides the discovery of the solution which Euclid gave, and which is substantially the one now commonly seen in textbooks. The problem was probably solved in some fashion before the time of Oenopides, however. The object of the problem is primarily to enable us to draw a line parallel to a given line.

Practically, the drawing of one line parallel to another is usually effected by means of a parallel ruler (see page 191), or by the use of draftsmen's triangles, as here shown, or even more commonly by the use of a T-square, such as is here seen. This illustration shows two T-squares used for drawing lines parallel to the sides of a board upon which the drawing paper is fastened.[71]

An ingenious instrument described by Baron Dupin is illustrated below.

To the bar _A_ is fastened the sliding check _B_. A movable check _D_ may be fastened by a screw _C_. A sharp point is fixed in _B_, so that as _D_ slides along the edge of a board, the point marks a line parallel to the edge. Moreover, _F_ and _G_ are two brass arms of equal length joined by a pointed screw _H_ that marks a line midway between _B_ and _D_. Furthermore, it is evident that _H_ will draw a line bisecting any irregular board if the checks _B_ and _D_ are kept in contact with the irregular edges.