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PROPOSITION 22. 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; namely, A and B greater than C; 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, and make DF equal to A, FG equal to B, and' GH equal to C.

[I. 3. From the centre F, at the distance FD, describe the circle

B DKL.

[Post. 3. From the centre G, at the distance GH, describe the circle HLK, cutting the former circle at K. Join KF, KG. The triangle KFG shall have its sides equal to the three straight lines A, B, C.

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

[Definition 15. But FD is equal to A.

[Construction. Therefore FK is equal to A.

[Axiom 1. Again, because the point G is the centre of the

circle HLK, GH is equal to GK.

[Definition 15. But GH is equal to C.

[Construction. Therefore GK is equal to C.

[Axiom 1. And FG is equal to B.

[Construction. Therefore the three straight lines KF, FG, GK are equal to the three A, B, C.

Wherefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines A, B, C. Q.E.F.

PROPOSITION 23. PROBLEM.

At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle: it is required to make at the given point A, in the given straight line AB, an angle equal to the given rectilineal angle DCE.

In CD, CE take any points D, É, and join DE. Make the triangle AFG the sides of which shall be equal to the three straight lines CD, DE, EC ; so that AF shall be equal to CD, AG to CE, and FG to DE. [I. 22. The angle FAG shall be equal to the angle DCE.

Because FA, AG are equal to DC, CE, each to each, and the base FG equal to the base DE; [Construction. therefore the angle FAG is equal to the angle DCE. [I. 8.

Wherefore at the given point A in the given straight line AB, the angle FAG has been made equal to the given rectilineal angle DCE. Q.E.F.

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PROPOSITION 24. 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, 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 the two sides AB, AC, equal to the two sides DE, DF, each to each, namely, AB to DE, and AC to DF, but the angle BAC greater than the angle EDF: the base BC shall be

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greater than the base EF.

Of the two sides DE, DF, let DE be the side which is not greater than the other. At the point D in the straight line DE,

G make the angle EDG equal to the angle BAC,

[I. 23. and make DG equal to AC or DF,

[I, 3. and join EG, GF. Because AB is equal to DE,

[Hypothesis. and AC to DG;

[Construction. the two sides BA, AC are equal to the two sides ED, DG, each to each ; and the angle BAC is equal to the angle EDG;. [Constr. therefore the base BC is equal to the base EG. [I. 4. And because DG is equal to DF,

[Construction. the angle DGF is equal to the angle DFG.

[I. 5. But the angle DGF is greater than the angle EGF. [Ax. 9. Therefore the angle DFG is greater than the angle EGF. Much more then is the angle EFG greater than the angle EGF.

[Axiom 9. And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle is subtended by the greater side,

[I. 19. therefore the side EG is greater than the side EF. But EG was shewn to be equal to BC; therefore BC is greater than EF.

Wherefore, if two triangles &c. Q.E.D.

PROPOSITION 25. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one

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greater than the base of the other, the angle contained by the sides of that 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 the two sides AB, AC equal to the two sides DE, DF, each to each, namely, AB to DE, and AC to DF, but the base BO'greater than the base EF: the angle BAC shall be greater than the angle EDF.

For if not, the angle BAC must be either equal to the angle EDF or than the angle EDF. But the angle BAC is not equal to the angle EDF, B for then the base BC would be equal to the base EF;

[I. 4. but it is not ;

[Hypothesis. therefore the angle BAC is not equal to the angle EDF. Neither is the angle BAC less than the angle EDF, for then the base BC would be less than the base EF; [I. 24. but it is not;

[Hypothesis. therefore the angle BAC is not less than the angle EDF. And it has been shewn that the angle BAC is not equal to the angle EDF. Therefore the angle BAC is greater than the angle EDF.

Wherefore, if two triangles &c. Q.E.D.

PROPOSITION 26. THEOREM. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, either the sides adjacent to the equal angles, or sides which are opposite to equal angles in each, then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other.

Let ABC, DEF be two triangles, which have the angles ABC, BCA equal to the angles DEF, EFD, each

to each, namely, ABC to DEF, and BCA to EFD; and let them have also one side equal to one side ; and first let those sides be equal which are adjacent to the equal angles in the two triangles, namely, BC to EF: the other sides shall be equal, each to each, namely, AB to DE, and AC to DF, and the third angle BAC equal to the third angle EDF.

For if AB be not equal to DE, one of them must be greater than the other. Let AB be the greater, and make BG

B equal to DE, [I. 3. and join GC.

Then because GB is equal to DE, [Construction. and BC to EF;

[Hypothesis. the two sides GB, BC are equal to the two sides DE, EF, each to each; and the angle GBC is equal to the angle DEF; [Hypothesis. therefore the triangle GBC is equal to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite;

[I. 4. therefore the angle GCB is equal to the angle DFE. But the angle DFE is equal to the angle ACB. (Hypothesis. Therefore the angle GCB is equal to the angle ACB, [Ax. 1. the less to the greater ; which is impossible. Therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF;

[Hypothesis. therefore the two sides AB, BC are equal to the two sides DE, EF, each to each ; and the angle ABC is equal to the angle DEF;[Hypothesis. therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF.

[I. 4.

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