The error is this, that the angles BEA and CE A are not the angles contained by the sides of the triangles which are equal, each to each. B E C The Fourth Proposition proves that two triangles are equal in every respect, if two sides of the one are equal to two sides of the other, each to each, and if the angles contained by those sides are also equal. Now, the sides (see the paragraph marked 3 of the above incorrect solution) are BA, AE of the triangle BA E, and C A, A E of the triangle CA E, and the angles contained by these sides are BAE and CAE respectively; and these angles are not given equal (the angles given equal are AEB and AEC), and therefore the triangles cannot be proved equal by the Fourth Proposition. In fact, the equality of B E and C E has to be proved by a subsequent Proposition, namely, the Twenty-sixth. This concludes the exercises on the Fourth Proposition; and we now pass on to the Fifth, a proposition which bears a name which makes it dreaded by beginners. Those, however, who have come thus far through this treatise need not fear it, for the simple reason that they have already done it. The foregoing Fifth and Sixth Exercises, with the appended deduction from them, is really the Fifth Proposition in a slightly disguised form. THE FIFTH PROPOSITION DISCUSSED. The enunciation of the Fifth Proposition may be put thus: It is given that the two sides of a triangle are equal, in other words, that the triangle is isosceles : It is required to prove that the angles at the base of the triangle are equal. And if the two equal sides of the isosceles triangle be produced: It is required to prove that the angles on the other side of the base are equal. Particular Enunciation. Let ABC be an isosceles triangle, having the side A B equal to the side A C: It is required to prove that the angle A B C is equal to the angle A C B. Also, if the equal sides AB, A C be produced beyond B and C, say to D the points D and E : : It is required to prove that the angle D B C is equal to the angle ECB. In order to prove this proposition, Euclid makes a couple of triangles-which he will be able to compare and prove equal in every respect by the Fourth Propositionin the following manner. In BD he takes any point F, and from A E, the greater [i.e. greater than A F1, he cuts off, by the Third Proposition, a part A G equal to A F, and he joins F C and B G. Now, if the learner turns back to Exercise V., he will recognise in the two triangles ADC and ABE a pair of triangles exactly corresponding to the two triangles A B G and ACF (page 79). C DE And these triangles can be proved equal in every respect, as the two triangles A B E and ADC were proved equal in Exercise V. This is the first part of Prop. V. Next, he can prove the triangles B C F and C BG (fig. p. 79) equal in every respect by Prop. IV., in the same с E manner as the triangles DBE and BDC were proved equal in every respect in Exercise VI. This is the second part of Prop. V. Lastly, he can deduce that the angle ABC is equal to the angle A CB (fig. p. 79), as ABD was proved equal to ADB in the deduction appended to the Fifth and Sixth Exercises. This is the third part of Prop. V. It will be a great assistance to the learner to notice particularly that the Fifth Proposition falls naturally into three parts, corresponding, respectively, to Exercises V. and VI., and to the deduction appended to those exercises. And now, with these hints, and references to these foregoing exercises, the learner is recommended boldly to try to write out the demonstration of the Fifth Proposition without professedly learning it. Let him carefully study Exercises V. and VI., and the appended deduction, also the hints given above; and then write out, if he can, the Fifth Proposition as an exercise on the Fourth. It is almost too much to hope that he will succeed in the endeavour, but it is worth trying, and if he succeeds it will be a great success. However, let him not be discouraged if he fails in his attempt to write out the demonstration without professedly learning the proposition. If he fails to attain this 'great success,' let him read carefully what follows, viz., the proposition written out in due form. If, after doing so, he finds that he is able to write out the demonstration, even this will be a very satisfactory amount of success, for many a poor fellow has spent weeks, if not months, over this proposition, and at the end of the time his mind has been in a state of utter confusion and bewilderment about it. THE FIFTH PROPOSITION WRITTEN OUT. PROPOSITION V. THEOREM. General Enunciation. It is given that a triangle is isosceles :It is required to prove that the angles at the base are equal. And if the equal sides be produced :— It is required to prove that the angles on the other side of the base are equal. Particular Enunciation. Let A B C be an isosceles triangle, having the side A B equal to A C: It is required to prove that the angle ABC is equal to the angle A CB. B E And if the equal sides A B and A C be produced to D and E: : It is required to prove that the angle D B C is equal to the angle ECB. B FA Construction. The sides A B and A C being produced to D and E, in BD take any point F, and from AE the greater cut off a part A G equal to A F the less, and join F C, B G. Demonstration. PART I. First, in the two triangles A B G and A CF: 1. A G equal to A F (construction). 2. A B equal to A C (hypothesis). 3. Wherefore the two sides FA, AC of the triangle FA C are equal to the two sides GA, A B of the triangle G A B, each to each. 4. And the included angle at A is common to the two triangles FAC and BAG. 5. Therefore, by Prop. IV., the two triangles F AC and BAG are equal in every respect. Wherefore the base BG is equal to the base FC, and the remaining angles, to which the equal sides are opposite, are equal, viz. the angle A B G to the angle AC F, and the angle A G B to the angle AF C. PART II. Secondly, in the two triangles FBC and GCB: 1. The side FC equal to the side B G. been proved so above. It has |