The Teaching of Geometry Part 14
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Let _ABC_ be the triangle, with _AB_ = _AC_. Conceive of this as two triangles; then _AB_ = _AC_, _AC_ = _AB_, and [L]_A_ is common; hence the [triangles]_ABC_ and _ACB_ are congruent, and [L]_B_ of the one equals [L]_C_ of the other.
This is a better plan than that followed by some textbook writers of imagining [triangle]_ABC_ taken up and laid down on _itself_. Even to lay it down on its "trace" is more objectionable than the plan of Pappus.
THEOREM. _If two angles of a triangle are equal, the sides opposite the equal angles are equal, and the triangle is isosceles._
The statement is, of course, tautological, the last five words being unnecessary from the mathematical standpoint, but of value at this stage of the student's progress as emphasizing the nature of the triangle.
Euclid stated the proposition thus, "If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another." He did not define "subtend," supposing such words to be already understood. This is the first case of a converse proposition in geometry. Heath distinguishes the logical from the geometric converse. The logical converse of Euclid I, 5, would be that "_some_ triangles with two angles equal are isosceles," while the geometric converse is the proposition as stated. Proclus called attention to two forms of converse (and in the course of the work, but not at this time, the teacher may have to do the same): (1) the complete converse, in which that which is given in one becomes that which is to be proved in the other, and vice versa, as in this and the preceding proposition; (2) the partial converse, in which two (or even more) things may be given, and a certain thing is to be proved, the converse being that one (or more) of the preceding things is now given, together with what was to be proved, and the other given thing is now to be proved. Symbolically, if it is given that _a_ = _b_ and _c_ = _d_, to prove that _x_ = _y_, the partial converse would have given _a_ = _b_ and _x_ = _y_, to prove that _c_ = _d_.
Several proofs for the proposition have been suggested, but a careful examination of all of them shows that the one given below is, all things considered, the best one for pupils beginning geometry and following the sequence laid down in this chapter. It has the sanction of some of the most eminent mathematicians, and while not as satisfactory in some respects as the _reductio ad absurdum_, mentioned below, it is more satisfactory in most particulars. The proof is as follows:
[Ill.u.s.tration:
=Given the triangle ABC, with the angle A equal to the angle B.=]
_To prove that_ _AC_ = _BC_.
=Proof.= Suppose the second triangle _A'B'C'_ to be an exact reproduction of the given triangle _ABC_.
Turn the triangle _A'B'C'_ over and place it upon _ABC_ so that _B'_ shall fall on _A_ and _A'_ shall fall on _B_.
Then _B'A'_ will coincide with _AB_.
Since [L]_A'_ = [L]_B'_, Given
and [L]_A_ = [L]_A'_, Hyp.
[therefore][L]_A_ = [L]_B'_.
[therefore]_B'C'_ will lie along _AC_.
Similarly, _A'C'_ will lie along _BC_.
Therefore _C'_ will fall on both _AC_ and _BC_, and hence at their intersection.
[therefore]_B'C'_ = _AC_.
But _B'C'_ was made equal to _BC_.
[therefore]_AC_ = _BC_. Q.E.D.
If the proposition should be postponed until after the one on the sum of the angles of a triangle, the proof would be simpler, but it is advantageous to couple it with its immediate predecessor. This simpler proof consists in bisecting the vertical angle, and then proving the two triangles congruent. Among the other proofs is that of the _reductio ad absurdum_, which the student might now meet, but which may better be postponed. The phrase _reductio ad absurdum_ seems likely to continue in spite of the efforts to find another one that is simpler. Such a proof is also called an indirect proof, but this term is not altogether satisfactory. Probably both names should be used, the Latin to explain the nature of the English. The Latin name is merely a translation of one of several Greek names used by Aristotle, a second being in English "proof by the impossible," and a third being "proof leading to the impossible." If teachers desire to introduce this form of proof here, it must be borne in mind that only one supposition can be made if such a proof is to be valid, for if two are made, then an absurd conclusion simply shows that either or both must be false, but we do not know which is false, or if only one is false.
THEOREM. _Two triangles are congruent if the three sides of the one are equal respectively to the three sides of the other._
It would be desirable to place this after the fourth proposition mentioned in this list if it could be done, so as to get the triangles in a group, but we need the fourth one for proving this, so that the arrangement cannot be made, at least with this method of proof.
This proposition is a "partial converse" of the second proposition in this list; for if the triangles are _ABC_ and _A'B'C'_, with sides _a_, _b_, _c_ and _a'_, _b'_, _c'_, then the second proposition a.s.serts that if _b_ = _b'_, _c_ = _c'_, and [L]_A_ = [L]_A'_, then _a_ = _a'_ and the triangles are congruent, while this proposition a.s.serts that if _a_ = _a'_, _b_ = _b'_, and _c_ = _c'_, then [L]_A_ = [L]_A'_ and the triangles are congruent.
The proposition was known at least as early as Aristotle's time. Euclid proved it by inserting a preliminary proposition to the effect that it is impossible to have on the same base _AB_ and the same side of it two different triangles _ABC_ and _ABC'_, with _AC_ = _AC'_, and _BC_ = _BC'_. The proof ordinarily given to-day, wherein the two triangles are constructed on opposite sides of the base, is due to Philo of Byzantium, who lived after Euclid's time but before the Christian era, and it is also given by Proclus. There are really three cases, if one wishes to be overparticular, corresponding to the three pairs of equal sides. But if we are allowed to take the longest side for the common base, only one case need be considered.
Of the applications of the proposition one of the most important relates to making a figure rigid by means of diagonals. For example, how many diagonals must be drawn in order to make a quadrilateral rigid? to make a pentagon rigid? a hexagon? a polygon of _n_ sides. In particular, the following questions may be asked of a cla.s.s:
[Ill.u.s.tration]
1. Three iron rods are hinged at the extremities, as shown in this figure. Is the figure rigid? Why?
2. Four iron rods are hinged, as shown in this figure. Is the figure rigid? If not, where would you put in the fifth rod to make it rigid? Prove that this would accomplish the result.
[Ill.u.s.tration]
Another interesting application relates to the most ancient form of leveling instrument known to us. This kind of level is pictured on very ancient monuments, and it is still used in many parts of the world.
Pupils in manual training may make such an instrument, and indeed one is easily made out of cardboard. If the plumb line pa.s.ses through the mid-point of the base, the two triangles are congruent and the plumb line is then perpendicular to the base. In other words, the base is level. With such simple primitive instruments, easily made by pupils, a good deal of practical mathematical work can be performed. The interesting old ill.u.s.tration here given shows how this form of level was used three hundred years ago.
[Ill.u.s.tration: EARLY METHODS OF LEVELING
Pomodoro's "La geometria prattica," Rome, 1624]
[Ill.u.s.tration]
Teachers who seek for geometric figures in practical mechanics will find this proposition ill.u.s.trated in the ordinary hoisting apparatus of the kind here shown. From the study of such forms and of simple roof and bridge trusses, a number of the usual properties of the isosceles triangle may be derived.
THEOREM. _The sum of two lines drawn from a given point to the extremities of a given line is greater than the sum of two other lines similarly drawn, but included by them._
It should be noted that the words "the extremities of" are necessary, for it is possible to draw from a certain point within a certain triangle two lines to the base such that their sum is greater than the sum of the other two sides.
[Ill.u.s.tration]
Thus, in the right triangle _ABC_ draw any line _CX_ from _C_ to the base. Make _XY_ = _AC_, and _CP_ = _PY_. Then it is easily shown that _PB_ + _PX_ > _CB_ + _CA_.
[Ill.u.s.tration]
It is interesting to a cla.s.s to have a teacher point out that, in this figure, _AP_ + _PB_ < _ac_="" +="" _cb_,="" and="" _ap'_="" +="" _p'b_="">< _ap_="" +="" _pb_,="" and="" that="" the="" nearer="" _p_="" gets="" to="" _ab_,="" the="" shorter="" _ap_="" +="" _pb_="" becomes,="" the="" limit="" being="" the="" line="" _ab_.="" from="" this="" we="" may="" _infer_="" (although="" we="" have="" not="" proved)="" that="" "a="" straight="" line="" (_ab_)="" is="" the="" shortest="" path="" between="" two="">
THEOREM. _Only one perpendicular can be drawn to a given line from a given external point._
THEOREM. _Two lines drawn from a point in a perpendicular to a given line, cutting off on the given line equal segments from the foot of the perpendicular, are equal and make equal angles with the perpendicular._
THEOREM. _Of two lines drawn from the same point in a perpendicular to a given line, cutting off on the line unequal segments from the foot of the perpendicular, the more remote is the greater._
THEOREM. _The perpendicular is the shortest line that can be drawn to a straight line from a given external point._
These four propositions, while known to the ancients and incidentally used, are not explicitly stated by Euclid. The reason seems to be that he interspersed his problems with his theorems, and in his Propositions 11 and 12, which treat of drawing a perpendicular to a line, the essential features of these theorems are proved. Further mention will be made of them when we come to consider the problems in question. Many textbook writers put the second and third of the four before the first, forgetting that the first is a.s.sumed in the other two, and hence should precede them.
THEOREM. _Two right triangles are congruent if the hypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other._
THEOREM. _Two right triangles are congruent if the hypotenuse and an adjacent angle of the one are equal respectively to the hypotenuse and an adjacent angle of the other._
As stated in the notes on the third proposition in this sequence, Euclid's c.u.mbersome Proposition 26 covers several cases, and these two among them. Of course this present proposition could more easily be proved after the one concerning the sum of the angles of a triangle, but the proof is so simple that it is better to leave the proposition here in connection with others concerning triangles.
The Teaching of Geometry Part 14
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The Teaching of Geometry Part 14 summary
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