Another demonstration of its first part is often adopted depending on the method of superposition, which is said to have been invented by Pappus, a geometer of the Alexandrian school who lived in the fourth century. Imagine the isosceles triangle ABC to be turned over and set down the reverse way. This will give, as it were, the A figure of two triangles ABC and ACB. B. the two. B In these AB = AC AC =AB. The angle at A is common to Therefore the triangles are equal in all respects (I. 4), and the angle ABC = ACB = ACB. Proposition VI. is what is called the converse of the first part of Proposition V., i.e. we are given what we were required to prove before, and we have to prove what we were then given. e.g. Proposition V. given-Two sides in a triangle equal; show that the two opposite angles are equal. Proposition VI. given-Two angles equal; show that the two opposite sides are equal. Euclid here, as in many other converse propositions, adopts the indirect method of proof. This, which is not so satisfactory to most minds as the direct proof, consists in showing that an absurdity or an impossibility, such as the contradiction of an established truth, necessarily follows from any supposition contrary to that in the enunciation. We have had an instance of the 'reductio ad absurdum' in I. 4 when it was shown that if the bases of the two triangles did not coincide, two straight lines would enclose a space, but as it was also made evident by superposition that they did coincide, this proof cannot be considered an example of indirect demonstration. The method will be best understood by an example, and as the argument is very simple we will take the proposition in Euclid's words. (See page 115.) B SUMMARY OF PROPOSITION VI., THEOREM 3. The angles at the base of a triangle being given equal, to show that the sides opposite to them are equal. A Cons.-Assuming AB to be greater than AC, make DB equal to it (I. 3), join DC. Proof (Indirect).-Show that ▲ DBC = ▲ ABC (I. 4). This is absurd (ax. 9). .. AB is not unequal to AC, i.e. AB= c AC. LECTURE V. PROPS. 7-8.-RESULTS OBTAINED FROM THE GROUP OF THEOREMS CONCLUDED. THE enunciation of Prop. VII. is negative, which is very unusual. It states that 'on the same base, and on the same side of it, there cannot be two triangles which have their sides which are terminated at one extremity of the base equal and likewise those which are terminated at the other.' An example will make the meaning of this clear: Let ABC, ABD be two triangles on the same base AB, if AC = AD it is impossible that BC should = BD; or, if BC = BD, AC cannot = AD. In drawing this figure you will observe that the triangles can be arranged in three different ways, each representing the conditions of the enunciation. (1) The vertex of each triangle may be outside of the other; (2) or, the vertex of one C D A A A B triangle may fall inside the other; or (3), it may B fall on a side of the other. In the last case it is plainly impossible that AD should be equal to AC, the part to the whole. D We must now consider the first two cases which are proved by the indirect method and by means of the 5th proposi B tion. In accordance with the indirect method we will assume that to be true which we wish to dis prove, and endeavour to deduce some absurd result from the assumption. Supposing AC and AD to be equal, and also CB and DB, by joining CD the vertices of the triangles we construct two isosceles triangles, with a common base CD. What do we know further about these triangles? Because they are isosceles the angles at their bases are equal, i.e. Angle (1) ACD = angle (2) ADC. (3) BCD Examining these four angles further we observe that two of them are parts of the other two. Hence : 1 is greater than 3, 4 is greater than 2. Combining this information, we may infer that— 1 is greater than 4 (since 3 = 4) | and that 4 is greater than 1 (since 2 = 1) which is absurd, and consequently the lines ending at A, and also those at B, are not equal. Or, the angles may be compared as follows: Since 1 = 2 and is greater than 3, 2 is greater than 3. Since 4 is greater than 2, much more is it greater than 3. But 4 is also equal to 3, which is absurd. The latter is the method adopted by Euclid. Case II.-By producing the sides of one of the isosceles triangles we are enabled to make a comparison of the angles similar to that in Case I. For angle (1) ECD = angle (2) FDC, because they are the angles on the other side of the base of the isosceles triangle ACD. Proceed as in Case I. Case III. has already been considered, A hence the theorem is completely demonstrated. E B SUMMARY OF PROPOSITION VII., THEOREM 4. On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated in one extremity of the base equal, and likewise those terminated at the other. Cons.-Cases I. and II. Join the vertices of the As. B E C F B Proof (Indirect).-Show that BDC is both equal to (I. 5) and greater than BCD. (In Case II., by means of the angles ECD and FDC on the other side of the base). Case III. is obvious. Proposition VIII. shows that two triangles are equal if the three sides of the one are equal to those of the other each to each. This is the second case of the equality of triangles demonstrated by Euclid. The proof depends upon the last proposition, which is introduced with the express object of leading to it. The method of superposition is employed as in I. 4. Apply one triangle to the other, and their bases being equal, will coincide. It follows that the other sides must coincide; for, if not, the condition exists which has just been shown to be impossible (I. 7). Therefore the triangles coincide and are equal in all respects (I. 4). Euclid only calls attention to the fact that their vertical angles are equal, because he wishes to draw attention to that particular circumstance in order to prove the next proposition. |