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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 1.

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 greater than the base EF.

B

Of the two sides DE, DF, let DE be the side 2. which is not greater than the other. At the point D in the straight line 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. 3.

Because AB is equal to DE [Hyp.], and AC

to DG [Const.]; 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 [Const.]; therefore the base BC is equal to the base EG. [I. 4.] And because DG is equal to DF (Const.), 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. [AX. 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 proved equal to BC; therefore BC is greater than EF.

Therefore if two triangles, &c.

Q.E.D.

“Of the two sides DE, DF, let DE be the side which is not greater than the other.” This condition is introduced in order to direct us on which side of the triangle EDF the construction is to be made. By starting from the side which is not greater than the other, the point G is made to fall above EF, otherwise there would be three cases to consider, viz.: 1. When G falls above EF, as in the proposition. 2. When G falls on EF, produced as fig. 1. 3. When G falls below EF, as in fig. 2.

D

2

The triangle EDG gives the corresponding shape of ABC in each case.

In the figure to the proposition the point G must lie in the circumference of a circle drawn from centre D, with radius DG, and as the angle EDG is greater than EDF, the point G must be above EP. Thus the two cases just mentioned have not to be considered by Euclid. This proposition should be compared with prop. 4, which it bears a relation similar to that which exists between props. 5 and 18. A similar remark applies to props. 25 and 8. See note to prop. 18. Ex. The two hands of a watch are together at 12 o'clock. Show

that in moving from this position the distance between thoir e.ctremities is continually increased until they come into one straight line.

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 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 BC greater than the base EF: the angle BAC shall be greater than the angle EDF.

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3.

For if not, the angle BAC is either equal to the

angle EDF or less than the angle EDF. But the angle BAC is not equal to the angle EDF, for then the base BC would be equal to the base EF (I. 4.] ; but it is not [Hyp.]; therefore the angle BAC is not equal to the angle EDF. Neither is the angle BAC less than the angle EDC, for then the base BC would be less than the base EF [I. 24]; but it is not [Hyp]; therefore the angle BAC is not less than the angle EDF. And it has been proved that the angle BAC is not equal to the angle EDF, Therefore the angle BAC is greater than the angle EDF.

Therefore, 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 1.

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. A

D

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For if AB be not equal to DE, one of them 2. must be greater than the other. Let AB be the greater, and make BG equal to DE, (I. 3.) and join GC.

Then because GB is equal to DE, [Con.) and 3.

BC to EF; (Hyp.] 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; [Hyp.] therefore the base GC is equal to the base DF, and the triangle GBC 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. [Hyp.] 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; [Hyp.] 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 angla DEF; [Hyp.] therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. [I. 4.]

Next, let sides which are opposite to equal 1.

angles in each triangle be equal to one another, namely, AB to DE: likewise in this case the other sides shall be equal, each to each, namely, BC to EF, and AC to DF; and also the third angle BAC equal to the third angle EDF.

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For if BC be not equal to EF, one of them 2.

must be greater than the other. Let BC be the greater, and make BH equal to EF (I. 3], and join AH.

Then because BH is equal to EF [Con.], and

AB to DE (Hyp.]; the two sides AB, BH are equal to the two sides DE, EF, each to each ; and the

3.

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