GEOMETRY 



2446 



GEOMETRY 



If two parallels are cut by a transversal, the 

 alternate interior angles are equal. 



Hypothesis: AB and CD are parallel lines, cut 

 by the transversal EF. 



Conclusion: Angle x Bangle y. 



-8 



pendicular to CD. " f A 



(It has been proved that "a straight line perpen- 

 dicular to one of two parallels is perpendicular to 

 the other." 



Apply the figure MKF to the figure MHE so 

 that the equal vertical angles at M shall coincide, 

 MK falling along MH, and MF along ME. (It 

 has been proved that all vertical angles are 

 equal.) 



Then F will fall on E, 



for MF equals ME by construction ; 

 and FK will fall along HE, 



for FK and EH are both perpendicular to HK. 

 (A previous theorem has proved that "from a 

 point without a line, only one perpendicular can 

 be drawn to the line.") 



Therefore the angles x and y coincide and are 

 equal, for by definition "the test of equality of 

 any two magnitudes is that they can be made to 

 coincide." Q.E.D. 



In an indirect proof something is assumed 

 as true, and known truths and demonstrated 

 propositions are built upon it until it is shown 

 that if the original assumption be not true the 

 result is an absurdity a self-evident falsity. 

 This method is known as the reductio ad 

 absurdum, or reduction to an absurdity. The 

 following demonstration gives a very simple 

 example of this interesting method: 



Two straight lines in the same plane, perpen- 

 dicular to the same straight line, are parallel. 



Hypothesis: AB and CD are two lines perpen- 

 dicular to the line EF. 



Conclusion: AB and CD are parallel. 



Proof. Could yc- 

 AB and CD, if 

 produced far 

 enough, meet at 

 some point, as x, 

 there would be 

 two perpendicu- 

 lars drawn from the point x to the line EF. But 

 this is impossible, for it has been proved that 

 "from a point without a line only one perpen- 

 dicular can be drawn to the line." 



Therefore AB and CD cannot meet. 



Therefore AB and CD are parallel. 



For by definition, "lines in the same plane which 



cannot meet, however far produced, are parallel." 



Q.E.D. 



It will be. evident from these two examples 

 that the order in which theorems are to be 

 demonstrated is fairly well determined, for 

 each demonstration must make use of only 



what has been proved before it cannot reach 

 forward to what is to come. In plane geom- 

 etry, the simplest form and therefore the most 

 interesting to most students, there are about 

 130 important propositions, but there are many 

 minor propositions which are no less interest- 

 ing, and many corollaries which are difficult to 

 prove, no matter how simple they may appear 

 to be. 



Some Famous Theorems. There are some 

 theorems which are particularly famous, either 

 because of some special difficulty, some neat- 

 ness in the demonstration, or some historical 

 interest. There is the pons asinorum, for in- 

 stance, as "asses' bridge." From time imme- 

 morial, it seems, dull scholars have found dif- 

 ficulty with this proposition, which Euclid 

 made the fifth in the first book of his Ele- 

 ments. Here it is, with its proof: 



In an isosceles triangle, the angles opposite the 

 equal sides are equal. 



Hypothesis: ABC is an isosceles triangle, hav- 

 ing AB equal to AC. 



Conclusion: Angle B = angle C. 



Proof. Draw * 



AD bisecting the 

 base BC. 

 The triangles 



ADB and ADC 



are equal ; 

 for AD=AD, by 

 identity, 

 DB = DC, by 

 construction, 

 and AB=AC, by hypothesis ; 

 and it has been proved that "two triangles are 

 equal if the three sides of one are equal respec- 

 tively to the three sides of the other." 

 Therefore angle B := angle C, 



for it has been proved that "homologous angles of 

 equal triangles are equal. Q.E.D. 



Then there is the so-called Pythagorean 

 Theorem, one of the most practical and useful 

 of all the propo- * 

 sitions of geom- 

 etry "The 

 square of the 

 hypotenuse of a 

 right-angled tri- 

 angle is equal to 

 the sum of the " 

 squares of the other two sides." Simple enough 

 it seems, when all the necessary preliminaries 

 are proved, but its demonstration, with that of 

 its necessary preliminaries, is one of the surest 

 proofs that Pythagoras was a brilliant reasoner. 



Hypothesis: BC is the hypotenuse of the right 

 triangle BAG. 



Conclusion: BC2 = AB2-f AC2. 



