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Required.–To prove base B D equal to base DC.

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II. Given.-Side A B equal to side A C.

Side A D equal to side A E.

B

Required.—To prove base BE equal to base DC. III. Figure ABCD is called a parallelogram. AD

is the diagonal. The opposite sides and angles of parallelograms are equal.

Prove.-That triangle A B D is equal to triangle

ACD.

THEOREM (Euclid I. 7). Assumed here as an axiom and explained. Proved on page 96. General Enunciation. On the same base, and on the same side of

it, there cannot be two triangles having the two sides terminated in one extremity of the base equal, and likewise those terminated in the other extremity

equal. Particular Enunciation. Given.—The two triangles C AB, DAB, on the

same base A B, and on the same side of it.

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Required.–To prove that the two sides CA, DA

cannot be equal, and have the sides CB, DB equal; or D B, CB be equal, and have D A,

CA equal. Observe: this proposition expresses the fact that

two triangles in the position of CAB, DAB, cannot have both pairs of sides equal. In

Fig. I neither pair of sides is equal. The triangles may be drawn with either pair of

sides equal, but then the other pair must be

unequal. Thus, in Fig. 2, sides CA and DA are drawn equal. Then it can be seen that CB and DB are unequal.

Fig.2

Fig.3

A

B

А.

Again, in Fig. 3, sides C B and D B are drawn

equal. And, as can be seen, CA and D A become unequal.

THEOREM (Euclid 1. 8). Repeat.-The enunciation of Euc. I. 7, and Axioms 8 and 8a. General Enunciation. If two triangles have two sides of the one

equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides, equal to them, of the other.

The same, in tabular form.

GIVEN.

REQUIRED.

Two triangles having

To prove(a) two sides of one That the angles con

equal to two sides of tained by the equal the other, each to each; sides are equal. (6) and the base of one equal to the base of the other.

NOTE.This proposition is the converse of Euc. I. 4. In that proposition we prove that, the angles being equal, the bases are equal. In this we prove that, the bases being equal, the angles are equal. Particular Enunciation. Given.-The two triangles A B C D E F having

the two sides BA, A C, equal to the two sides ED, DF, each to each ; and the base BC equal to the base E F.

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Required.—To prove that angle B A C is equal

to angle EDF. Proof. Let the triangle A B C be applied to the triangle

DEF, so that (a) the point B (Fig. 1) may be on the point E, and the straight line B Con EF. Then

(6) The point C shall coincide with the

point F, Because B C is equal to E F (Axiom 8a). (If B C were less than EF, C would fall between E and F; if B C were longer than EF, C would fall beyond F; but since B C is equal

to EF, C falls on F.) The point B coinciding with E (a), and the point

C coinciding with F (6). (c) The two straight lines BA, AC, will

coincide with the two straight lines ED,

DF. For if BA, A C do not coincide with ED, DF

but have a different situation as E G, GF (Fig. 2), then

D

Fig. 2.

On the same base E F, and on the same side of it,

we should have two triangles, DEF, GEF, having the two sides D E, G E equal, and also

the two sides D F, G F equal. But this is impossible (Euc. I. 7). ...BA, AC must coincide with ED, DF. and the angle BAC coincides with the

angle EDF, and is equal to it. Q. E, D. EXERCISES.-(The same method of reasoning to be employed as in the above theorem.) I. Given.—Side A C equal to side CB.

Base A D equa to base D B.

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