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Пур.

Proof. Because, if BC is not equal to EF, one of them must be

the greater. If possible, let BC be the greater, and from BC cut off BH equal to EF,

Prop. 3
Join AH,
Because in the two triangles ABH, DEF
AB is equal to DE,

Hyp. Because and BH is equal to EF,

Constr. and the contained angle ABH is equal to the contained

angle DEF; therefore by Prop. 4 the triangles ABH and DEF are equal

in all respects,
therefore the angle AHB is equal to the angle DFE.

But the angle ACB is equal to the angle DFE,

therefore the angle AHB is equal to the angle ACB: that is, the exterior angle AHB is equal to its interior opposite angle ACH, which is impossible.

Prop. 16 Therefore BC is not unequal to EF, that is BC is equal to EF.

Again, because in the two triangles ABC, DEF
AB is equal to DE,

Нур. Because and BC is equal to EF,

Proved and the contained angle ABC is equal to the contained angle DEF;

Нур. therefore by Prop. 4 the triangle ABC is equal to the

triangle DEF in all respects : that is, AC is equal to DF; the angle BAC is equal to the angle EDF; and the triangle ABC is equal to the triangle DEF in area.

Q. E. D.

EXERCISE. Write out the proof of Prop. 26 assuming that AC and DF are the only sides given as equal.

EUC.

E

NOTES ON PROPOSITIONS 21—26.

Proposition 21.- This proposition is important, and is often useful for proving riders; but it is itself really a rider, the first part depending on Prop. 20, and the second part on Prop. 16.

Proposition 22.-Any two of the three straight lines A, B, C must be greater than the third, because Prop. 20 proves that any two sides of a triangle must be greater than the third side, and therefore we could not describe a triangle with its sides equal to A, B, C if any two of them were not greater than the third.

The method of constructing the triangle KFG in Prop. 22 is really the same as the method used in Prop. 1 to construct the triangle ABC. Euclid's method of describing a triangle with its sides of given lengths is to take a line FG equal to one of the sides ; then, in order to find the vertex of the triangle, to describe two circles from Fand G as centres and with radii equal in length to the other two sides of the triangle; the intersection of the circles gives the vertex. It should be noticed that

(1) Two triangles might be drawn in each case; because the circles intersect in two points, one on each side of the base.

(2) If the required triangle is isosceles, the two lines A and C and the radii will be equal.

(3) If the required triangle is equilateral, the three lines A, B, C will be equal, and the radii will be equal to the base.

(4) If the triangle has to be described on a given line as base, that line must be used instead of the line FG.

Proposition 23.-This is an easy rider, the construction depending on Prop. 22, and the proof on Prop. 8.

Proposition 24.DE is assumed to be not greater than DF in order to make sure that the point F will fall below EG. If the point F fell either above or on EG, Euclid's proof would have to be altered.

This proposition should be compared with Prop. 4. In Prop. 4 the base BC is proved equal to the base EF, and in Prop. 24 the base BC is proved greater than the base EF,

The proof of Prop. 24 is divided into two parts. First, the base BC is proved equal to the base EG by Prop. 4; and then the angle EFG having been proved by the “Much more then” method of proof, as in Prop. 7, to be greater than the angle EGF, it follows by Prop. 19 that EG is greater than EF. Prop. 24 may therefore be said to be a rider on Props. 4 and 19.

Proposition 25.—This proposition is the converse of Prop. 24. In Prop. 24 we take for granted that the angle BAC is greater than the angle EDF, and we prove that the base BC is greater than the base EF; conversely in Prop. 25 we take for granted the base and prove the angle. Proposition 25 should be compared with Prop. 8. In Prop. 8 the angle BAC is proved equal to the angle EDF, and in Prop. 25 the angle BAC is proved greater than the angle EDF. Prop. 25 is proved like Prop. 19 by the method of exhaustion. The two impossible cases depend on Props. 4 and 24.

Proposition 26.—This really contains two propositions, which we give as Case I. and Case II. Both are proved by a Reductio ad Absurdum, in Case I. the Ninth Axiom, and in Case II. Prop. 16 being used. The Proofs of both cases are made up of two simple applications of Prop. 4.

It might be supposed that a Case III. should be mentioned, namely, when the side AC is given equal to the side DF. But this is really the same as Case II., the equal sides being opposite to equal angles.

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EASY EXERCISES ON PROPOSITIONS 21—26. 1. Take any point 0 within the triangle ABC and join 0A, OB, OC. Prove that 0A, OB, OC are together less than AB, BC, CA.

2. Describe an angle double of a given angle.

3. Show how to construct a quadrilateral figure equal in all respects to a given quadrilateral figure.

4. In the fig. of Prop. 21 join AD and produce it to meet the base BC in F; by this means prove that the angle BDC is greater than the angle BAC.

5. Take any finite straight line AB and any two rectilineal angles C and D; show how to construct on AB a triangle having two of its angles equal to C and D. When is this impossible ?

6. In the fig. of Prop. 24 bisect the angle FDG by the straight line DH, which meets EG in H, and join FH. Prove that FH is equal to HG, and hence prove that BC is greater than EF.

7. In the triangle ABC join the vertex A to M the middle point of the base BC, and prove that the angle AMB is greater than, equal to, or less than the angle AMC according as AB is greater than, equal to, or less than AC.

8. ABC is an isosceles triangle having the side AB equal to the side AC. Bisect each of the angles ABC, ACB by the straight lines BY, CZ which meet AC and AB in Y and Z respectively. Prove that the triangles YBC, ZCB are equal in all respects.

9. Prove by Reductio ad Absurdum, as in Prop. 26, that if two right-angled triangles have their hypotenuses equal and one side equal to one side, the triangles are equal in all respects.

SUMMARY OF THE RESULTS ARRIVED AT IN EUCLID I. 1—26.

The first 26 Propositions in Book I. deal with Lines, Angles, and Triangles. Lines.-(a) Props. 2, 3, 10, 11, 12 are Problems on the drawing

and bisecting of lines. (6) Props. 13, 14, 15 are Theorems proving that if

two lines cut each other the angles on one side are together equal to two right angles, and that

the vertically opposite angles are equal. Angles.- Props. 9 and 23 are Problems on bisecting or describing

an angle. Triangles.-(a) Props. 1 and 22 are Problems on describing a

triangle. (6) Props. 5 and 6 are Theorems, dealing with an

isosceles triangle. (c) Props. 16, 17, 18, 19, 20, 21 are Theorems about

a single triangle, and the chief results arrived

at are that(1) The exterior angle is greater than either interior

opposite angle. (2) Any two angles are less than two right angles. (3) The greater side and greater angle are opposite each

other. (4) Any two sides of a triangle are greater than the third.

(d) Props. 4, 8, 26 are Theorems dealing with the

equality of two triangles; and 24, 25 are Theorems dealing with the inequality of two

triangles. Of all the Propositions in this part of Euclid, Props. 4, 8, 26 are the most important. In order to prove two triangles equal in all respects, we must have given at least three parts of one triangle equal to three parts of the other; but even then in two cases the triangles may not be equal in all respects, as the following diagrams will show.

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CASE VI.-Two sides and an angle not included by them equal

Triangles are not necessarily
equal, for there

may

be two positions as DG and DF for

B the shorter side.

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In order therefore to prove two triangles equal in all respects recourse must be had to one of the first four of these Cases,

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