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It has been suggested to demonstrate I. 5 by superposition. Conceive the isosceles triangle ABC to be taken up, and then replaced so that AB falls on the old position of AC, and AC falls on the old position of AB. Thus, in the manner of I. 4, we can shew that the angle ABC is equal to the angle ACB.

I. 6 is the converse of part of I. 5. One proposition is said to be the converse of another when the conclusion of each is the hypothesis of the other. Thus in I. 5 the hypothesis is the equality of the sides, and one conclusion is the equality of the angles; in I. 6 the hypothesis is the equality of the angles and the conclusion is the equality of the sides. When there is more than one hypothesis or more than one conclusion to a proposition, we can form more than one converse proposition. For example, as another converse of I. 5 we have the following: if the angles formed by the base of a triangle and the sides produced be equal, the sides of the triangle are equal; this proposition is true and will serve as an exercise for the student.

The converse of a true proposition is not necessarily true; the student however will see, as he proceeds, that Euclid shews that the converses of many geometrical propositions are true.

I. 6 is an example of the indirect mode of demonstration, in which a result is established by shewing that some absurdity follows from supposing the required result to be untrue. Hence this mode of demonstration is called the reductio ad absurdum. Indirect demonstrations are often less esteemed than direct demonstrations; they are said to shew that a theorem is true rather than to shew why it is true. Euclid uses the reductio ad absur. dum chiefly when he is demonstrating the converse of some former theorem; see I. 14, 19, 25, 40.

Some remarks on indirect demonstration by Professor Sylvester, Professor De Morgan, and Dr Adamson will be found in the volumes of the Philosophical Magazine for 1852 and 1853.

I. 6 is not required by Euclid before he reaches II. 4; so that I. 6 might be removed from its present place and demonstrated hereafter in other ways if we please. For example, I. 6 might be placed after I. 18 and demonstrated thus. Let the angle ABC be equal to the angle ACB: then the side AB shall be equal to the side AC. For if not, one of them must be greater than the other; suppose AB greater than AC. Then the angle ACB is greater But this is impossible, because

than the angle ABC, by I. 18.

the angle ACB is equal to the angle ABC, by hypothesis.

Or I. 6 might be placed after I. 26 and demonstrated thus. Bisect the angle BAC by a straight line meeting the base at D. Then the triangles ABD and ACD are equal in all respects, by I. 26. I. 7 is only required in order to lead to I. 8. be superseded by another demonstration of I. 8, recommended by many writers.

The two might

which has been

Let ABC, DEF be two triangles, having the sides AB, AC equal to the sides DE, DF, each to each, and the base BC equal to the base EF: the angle BAC shall be equal to the angle EDF.

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For, let the triangle DEF be applied to the triangle ABC, so that the bases may coincide, the equal sides be conterminous, and the vertices fall on opposite sides of the base. Let GBC represent the triangle DEF thus applied, so that G corresponds to D. Join AG. Since, by hypothesis, BA is equal to BG, the angle BAG is equal to the angle BGA, by I. 5. In the same manner the angle CAG is equal to the angle CGA. Therefore the whole angle BAC is equal to the whole angle BGC, that is, to the angle Edf.

There are two other cases; for the straight line AG may pass through B or C, or it may fall outside BC: these cases may be treated in the same manner as that which we have considered.

I. 8. It may be observed that the two triangles in I. 8 are equal in all respects; Euclid however does not assert more than the equality of the angles opposite to the bases, and when he requires more than this result he obtains it by using I. 4.

I. 9. Here the equilateral triangle DEF is to be described on the side remote from A, because if it were described on the same side, its vertex, F, might coincide with A, and then the construction would fail.

I. 11. The corollary was added by Simson. It is liable to serious objection. For we do not know how the perpendicular BE is to be drawn. If we are to use I. II we must produce AB, and then we must assume that there is only one way of producing AB, for otherwise we shall not know that there is only one perpendicular; and thus we assume what we have to demonstrate.

Simson's corollary might come after I. 13 and be demonstrated thus. If possible let the two straight lines ABC, ABD have the segment AB common to both. From the point B draw any straight line BE. Then the angles ABE and EBC are equal to two right angles, by I. 13, and the angles ABE and EBD are also equal to two right angles, by I. 13. Therefore the angles ABE and EBC are equal to the angles ABE and EBD. Therefore the angle EBC is equal to the angle EBD; which is absurd.

But if the question whether two straight lines can have a common segment is to be considered at all in the Elements, it might occur at an earlier place than Simson has assigned to it. For example, in the figure to I. 5, if two straight lines could have a common segment AB, and then separate at B, we should obtain two different angles formed on the other side of BC by these produced parts, and each of them would be equal to the angle BCG. The opinion has been maintained that even in I. I, it is tacitly assumed that the straight lines AC and BC cannot have a common segment at C where they meet; see Camerer's Euclid, pages 30 and 36.

Simson never formally refers to his corollary until XI. 1. The corollary should be omitted, and the tenth axiom should be extended so as to amount to the following; if two straight lines coincide in two points they must coincide both beyond and between those points.

I. 12. Here the straight line is said to be of unlimited length, in order that we may ensure that it shall meet the circle.

Euclid distinguishes between the terms at right angles and perpendicular. He uses the term at right angles when the straight line is drawn from a point in another, as in I. II; and he uses the term perpendicular when the straight line is drawn from a point without another, as in I. 12. This distinction however is often disregarded by modern writers.

I. 14. Here Euclid first requires his eleventh axiom.

For

in the demonstration we have the angles ABC and ABE equal to two right angles, and also the angles ABC and ABD equal to two right angles; and then the former two right angles are equal to the latter two right angles by the aid of the eleventh axiom. Many modern editions of Euclid however refer only to the first axiom, as if that alone were sufficient; a similar remark applies to the demonstrations of I. 15, and I. 28. In these cases we have omitted the reference purposely, in order to avoid perplexing a beginner; but when his attention is thus drawn to the circumstance he will see that the first and eleventh axioms are both used.

We may observe that errors, in the references with respect to the eleventh axiom, occur in other places in many modern editions of Euclid. Thus for example in III. 1, at the step "therefore the angle FDB is equal to the angle GDB," a reference is given to the first axiom instead of to the eleventh.

There seems no objection on Euclid's principles to the following demonstration of his eleventh axiom.

Let AB be at right angles to DAC at the point A, and EF at right angles to HEG at the point E: then shall the angles BAC and FEG be equal.

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Take any length AC, and make AD, EH, EG all equal to AC. Now apply HEG to DAC, so that H may be on D, and HG on DC, and B and F on the same side of DC; then G will coincide with C, and E with A. Also EF shall coincide with AB; for if not, suppose, if possible, that it takes a different position as AK. Then the angle DAK is equal to the angle HEF, and the angle CAK to the angle GEF; but the angles HEF and GEF are equal, by hypothesis; therefore the angles DAK and CAK are equal. But the angles DAB and CAB are also equal, by hypothesis; and the angle CAB is greater than the angle CAK; there

fore the angle DAB is greater than the angle CAK. Much more then is the angle DAK greater than the angle CAK. But the angle DAK was shewn to be equal to the angle CAK; which is absurd. Therefore EF must coincide with AB; and therefore the angle FEG coincides with the angle BAC, and is equal to it.

I. 18, I. 19. In order to assist the student in remembering which of these two propositions is demonstrated directly and which indirectly, it may be observed that the order is similar to that in I. 5 and I. 6.

I. 20. "Proclus, in his commentary, relates, that the Epicureans derided Prop. 20, as being manifest even to asses, and needing no demonstration; and his answer is, that though the truth of it be manifest to our senses, yet it is science which must give the reason why two sides of a triangle are greater than the third: but the right answer to this objection against this and the 21st, and some other plain propositions, is, that the number of axioms ought not to be increased without necessity, as it must be if these propositions be not demonstrated." Simson.

I. 21.

Here it must be carefully observed that the two straight lines are to be drawn from the ends of the side of the triangle. If this condition be omitted the two straight lines will not necessarily be less than two sides of the triangle.

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I. 22. "Some authors blame Euclid because he does not demonstrate that the two circles made use of in the construction of this problem must cut one another: but this is very plain from the determination he has given, namely, that any two of the straight lines DF, FG, GH, must be greater than the third. For who is so dull, though only beginning to learn the Elements, as not to perceive that the circle described from the centre F, at the distance FD, must meet FH betwixt F and H, because FD is less than FH; and that for the like reason, the circle described from the centre G, at the distance GH...must mest DG betwixt D and G; and that these circles must meet one another, because FD and GH are together greater than FG?" Simson.

The condition that B and C are greater than A, ensures that the circle described from the centre G shall not fall entirely within the circle described from the centre F; the condition that A and B are greater than C, ensures that the circle described

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