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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 one greater than the base of the other ; then the angle contained by the sides of that which has the greater base, shall be greater than the angle contained by the corresponding sides of the other.

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Let ABC, DEF be two triangles in which

the side BA is equal to the side ED, and the side AC is equal to the side DF,

but the base BC is greater than the base EF. Then shall the angle BAC be greater than the angle EDF.

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

Hyp Therefore the angle BAC is neither equal to, nor less than

the angle EDF; that is, the angle BAC is greater than the angle EDF. Q.E.D.

EXERCISE.

In a triangle ABC, the vertex A is joined to X, the middle point of the base BC; shew that the angle AXB is obtuse or acute, according as AB is greater or less than AC.

PROPOSITION 26. THEOREM.

If two triangles have two angles of the one equal to two angles of the other, each to each, and a side of one equal to a side of the other, these sides being either adjacent to the equal angles, or opposite to equal angles in each , then shall the triangles be equal in all respects.

CASE I. When the equal sides are adjacent to the equal angles in the two triangles.

D

E

I. 3.

B

F
Let ABC, DEF be two triangles, in which

the angle ABC is equal to the angle DEF,
and the angle ACB is equal to the angle DFE,

and the side BC is equal to the side EF. Then shall the triangle ABC be equal to the triangle DEF in all respects ; that is, AB shall be equal to DE, and AC to DF,

and the angle BAC shall be equal to the angle EDF. For if AB be not equal to DE, one must be greater than the other. If possible, let AB be greater than DE. Construction. From BA cut off BG equal to ED,

and join GC.
Proof. Then in the two triangles GBC, DEF,
GB is equal to DE,

Constr. and BC is equal to EF,

Hyp. Because

also the contained angle GBC is equal to the contained angle DEF;

Нур. therefore the triangle GBC is equal to the triangle DEF in all respects;

I. 4, so that the angle GCB is equal to the angle DFE.

But the angle ACB is equal to the angle DFE; Hyp. therefore also the angle GCB is equal to the angle ACB; Ax.1.

the part equal to the whole, which is impossible.

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Therefore AB is not unequal to 'DE;

that is, AB is equal to DE. Hence in the triangles ABC, DEF, AB is equal to DE,

Proved. and BC is equal to EF;

Hyp. Because

also the contained angle ABC is equal to the contained angle DEF;

Нур. therefore the triangle ABC is equal to the triangle DEF in all respects :

I. 4. so that the side AC is equal to the side DF; and the angle BAC is equal to the angle EDF.

Q.E.D.

CASE II. When the equal sides are opposite to equal angles in the two triangles.

[blocks in formation]

Let ABC, DEF be two triangles, in which

the angle ABC is equal to the angle DEF, and the angle ACB is equal to the angle DFE,

and the side AB is equal to the side DE. Then the triangle ABC shall be equal to the triangle DEF in all respects ;

namely, BC shall be equal to EF,

and AC shall be equal to DF,
and the angle BAC shall be equal to the angle EDF.

I. 3.

For if BC be not equal to EF, one must be greater than the other. If possible, let BC be greater than EF. Construction. From BC cut off BH equal to EF,

and join AH. i Proof. Then in the triangles ABH, DEF, AB is equal to DE,

Нур. and BH is equal to EF,

Constr. Because

also the contained angle ABH is equal to the contained angle DEF;

Нур. therefore the triangle ABH is equal to the triangle DEF in all respects;

I. 4. so that the angle AHB is equal to the angle DFE.

But the angle DFE is equal to the angle ACB; Hyp. therefore the angle AHB is equal to the angle ACB; Ax. 1. that is, an exterior angle of the triangle ACH is equal to an interior opposite angle; which is impossible.

Therefore BC is not unequal to EF,

that is, BC is equal to EF. Hence in the triangles ABC, DEF, AB is equal to DE,

Hyp. and BC is equal to EF ;

Proved. Because

also the contained angle ABC is equal to the contained angle DEF;

Hyp. therefore the triangle ABC is equal to the triangle DEF in all respects;

I. 4. so that the side AC is equal to the side DF, and the angle BAC is equal to the angle EDF.

Q.E.D.

I. 16.

COROLLARY. In both cases of this Proposition it is seen that the triangles may be made to coincide with one another; and they are therefore equal in area.

ON THE IDENTICAL EQUALITY OF TRIANGLES.

Three cases have been already dealt with in Propositions 4, 8, and 26, the results of which may be summarized as follows :

Two triangles are equal in all respects when the following three parts in each are severally equal : 1. Two sides, and the included angle.

Prop. 4. 2. The three sides.

Prop. 8, Cor. 3. (athe adjacent

6) Two angles, and a side opposite one of them.} Prop. 26.

Two triangles are not, however, necessarily equal in all respects when any three parts of one are equal to the corresponding parts of the other. For example

(i) When the three angles of one are equal to the three angles of the other, each to each, the adjoining diagram shews that the triangles need not be equal in all respects.

(ii) When two sides and one angle in one are equal to two sides and one angle in the other, the given angles being opposite to equal sides, the diagram shews that the triangles need not be equal in all respects.

For it will be seen that if
AB=DE, and AC=DF, and

B
с E

F' F the_angle ABC = the angle DEF, then the shorter of the given sides in the triangle DEF may lie in either of the positions DF or DF'.

In cases (i) and (ii) a further condition must be given before we can prove that the two triangles are identically equal.

=

We observe that in each of the three cases in which two triangles have been proved equal in all respects, namely in Propositions 4, 8, 26, it is shewn that the triangles may be made to coincide with one another; so that they are equal in area. Euclid however restricted himself to the use of Prop. 4, when he required to deduce the equality in area of two triangles from the equality of certain of their parts. This restriction is now generally abandoned.

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