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60. PROPOSITION XXII.- PROBLEM.

To make a triangle, of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third.

Let A, B, C be the three given straight lines, of which

any two whatever are greater than the third ; (I. 20) namely, A and B greater than 0, A and C greater than B,

and B and C greater than A. It is required to make a triangle of which the sides shall be

equal to A, B, C, each to each. Take a straight line DE, terminated at the point D, but unlimited towards E.

Make DF equal to A, FG equal to B, and GH equal to C;(I.3) at the distance FD from the centre F describe the circle DKL; at the distance GH from the centre G describe the circle HLK; and join KF, KG. Then the triangle KFG shall have its sides equal to the three

given straight lines.

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Because the point F is the centre of the circle DKL, therefore FD is equal to FK;

(Def. 15) but FD is equal to the straight line A;

(Hyp.) therefore FK is equal to A.

(Ax. 1) Again, because G is the centre of the circle HLK, therefore GH is equal to GK; but GH is equal to C;

therefore, also, GK is equal to C.

And FG was made equal to B. Therefore the three straight lines KF, FG, GK are respec

tively equal to the three A, B, 0;

that is, the triangle KFG has been made having its three sides equal to the three straight lines A, B, C, each to each.

Q. E. F. EXERCISE XXVI. 1. Construct the figure for the case in which the straight lines A and B are equal. 2. Also when A and Care equal.

3. Give construction and proof when B and C are equal.

4. Suppose a triangle ABC given; show how to make a triangle whose sides shall be respectively equal to the sides of ABC.

5. Show that the triangle constructed has also its angles respectively equal to those of ABC.

6. Given two straight lines A and B; show how to construct a rhombus having its diagonal equal to A, and each side equal to B.

61. PROPOSITION XXIII.-PROBLEM. To draw a straight line from a given point in a given straight line, which shall make an angle with it equal to a given angle.

Let AB be the given straight line, A the given point in it, and DCE the given angle.

It is required to draw from A a straight line which shall make an angle with AB equal to DCE.

In CD, CE take any points D, E, and join DE; make the triangle AFG, the sides of which shall be equal to the three straight lines CE, CD, DE,

(I. 22) so that AG is equal to CE, AF to CD, and FG to DE.

Then the angle FAG shall be equal to the angle DCE.

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F B

Because FA, AG are equal to DC, CE, each to each, and the base FG is equal to the base DE ;

therefore the angle FAG is equal to the angle DCE. (I. 8)

Wherefore, at a given point A in the given line AB, the angle FAG is made equal to the given angle DCE.

Q. E. F.

EXERCISE XXVII.

For the figure, see that of the next Article, in which it is given that

AC and AD are equal. 1. Which is the greater of the angles DCA, DCB? Quote the appropriate axiom.

2. Which is the greater of ADC and CDB ?
3. Which is the greater, ACD or BDC?
4. Show that the angle BDC is greater than the angle BCD.

5. Infer from the previous result which is the greater, BC or BD in the triangle BCD.

62. ON LEMMAS. Def.-A Lemma is a proposition whose principal object is to assist in, or to simplify, the demonstration of some other proposition; which, in general, it immediately precedes.

Proposition VII. may be considered a mere lemma to VIII. ; since, but for its use in the latter proposition, we should probably not find it in Euclid's Elements at all. But it would not be correct to say, for example, of Prop. XXIII., that it is a lemma to XXIV.; because, although required in that proposition, it is also of frequent use elsewhere, and is, therefore, an independent proposition.

Prop. XXIV., as given in Euclid's Elements, really is in need of a lemma in order to clearly account for the method of its construction. One which will answer this purpose has been worked out in substance in the course of the preceding Exercise. It will now be formally given. Unless this lemma, or its equivalent, be taken into account, there should be three figures employed in the demonstration of Prop. XXIV.; in one of which F is upon EG, and in another above EG.

LEMMA TO PROP. XXIV. If two triangles upon one base, and on one side of it, have the sides terminated in one extremity of the base equal, and each not less than the base, the vertex of either triangle will be without the other.

Given, on the same base AB and on the same side of it, the triangles ABC, ABD, placed as in annexed diagram, and which have AC equal to AD and each not less than AB;

then D and A will be on opposite sides of BC. B

Proof.
With centre A and radius AC describe a circle.
Since AC, AD are equal, this will pass through D.

Since AB is either equal to the radius or less than it,
B will either be on the circumference or within the circle.
Hence triangle BCD has its base BC within the circumference ;
and its vertex D on the circumference.

Therefore D is on the other side of BC from the centre A.

63. PROPOSITION XXIV.-THEOREM.

.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them of the other; then the base of that which has the greater angle shall be greater than the base of the other.

Let ABC, DEF be two triangles,
which have AB equal to DE, and AC to DF;
but the angle BAC greater than the angle EDF.

Then the base BC shall be greater than the base EF.
Of the two sides DE, DF, let DE be not the greater.

At the point D, in the straight line DE, make the angle EDG equal to the angle BAC; (I. 23) make DG equal to DF or AC (I. 3); and join EG, GF.

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Then, in the triangles ABC and DEG, because AB is equal to DE, and AC to DG, and the angle BAC to the angle EDG;

therefore the base BC is equal to the base EG. (I. 4)

And because, in the triangle DFG, DG is equal to DF, therefore the angle DFG is equal to the angle DGF; (I. 5) but DGF is greater than the angle EGF,

(Ax. 9) therefore DFG is also greater than EGF;

even more, then, is the angle EFG greater than EGF.
Now these are both angles of the triangle EFG;

and the greater angle is subtended by the greater side ; (I. 19)

therefore the side EG is greater than the side EF. But BC was proved equal to EG, therefore BC is greater than EF. Wherefore, if two triangles, &c.

Q. E. D.

64. PROPOSITION XXV.—THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other ; then the angle contained by the sides of the one which has the greater base, shall be greater than the angle contained by the sides equal to them of the other.

Let ABC, DEF be two triangles, which have AB equal to DE, and AC to DF; but the base BC greater than the base EF.

Then the angle BAC shall be greater than the angle EDF.

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For, if the angle BAC be not greater than the angle EDF, it must either be equal to it or less than it.

If BAC were equal to EDF, the base BC would be equal to the base EF;

(I. 4) but it is not equal;

(Hyp.) therefore BAC is not equal to EDF.

Again, if BAC were less than EDF, then BC would be less than EF;

(I. 24) but it is not less ;

(Hyp.) therefore BAC is not less than EDF. And it has been shown that BAC is not equal to EDF; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c.

Q. E. D.

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