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Required. To prove base BD equal to base DC.
II. Given.-Side A B equal to side A C.
Side A D equal to side A E.
Required. To prove base BE equal to base DC. III. Figure ABCD is called a parallelogram. AÐ
is the diagonal. The opposite sides and angles of parallelograms are equal.
Prove. That triangle ABD is equal to triangle ACD.
THEOREM (Euclid I. 7).
Assumed here as an axiom and explained. Proved on page 96.
On the same base, and on the same side of it, there cannot be two triangles having the two sides terminated in one tremity of the base equal, and likewise those terminated in the other extremity equal.
Given. The two triangles CAB, DA B, on the same base AB, and on the same side of it.
Required. To prove that the two sides CA, DA cannot be equal, and have the sides CB, D B equal; or DB, CB be equal, and have DA, CA equal.
Observe this proposition expresses the fact that two triangles in the position of CA B, DA B,
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.
Again, in Fig. 3, sides CB and DB are drawn equal.
And, as can be seen, CA and DA become unequal.
THEOREM (Euclid I. 8).
Repeat. The enunciation of Euc. I. 7, and Axioms 8 and 8a.
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.
NOTE.—This proposition is the converse of Euc. I. 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.
Given. The two triangles A B C, D E F having the two sides B A, A C, equal to the two sides ED, DF, each to each; and the base BC equal to the base EF
Required. To prove that angle B A C is equal to angle EDĒ.
Let the triangle ABC 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 C on EF. Then
(b) 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 E F, C would fall between E and F; if BC were longer than EF, C would fall beyond F; but since BC is equal to EF, C falls on F)
The point B coinciding with E (a), and the point C coinciding with F(b).
(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
On the same base E F, and on the same side of it, we should have two triangles, DEF, GEF, having the two sides DE, GE equal, and also the two sides DF, G F equal.
But this is impossible (Euc. I. 7). ... BA, AC must coincide with E D, 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 DB.