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(c). In the triangle AEB, the two sides BA, AE, are greater than BE.
To each of these equals add E C.
(BA, A E, and E C make B A and A C.)
(d) BA and AC are greater than BE and EC (c).
Again, by Euc. I. 20,
(e) In the triangle EDC, the two sides CE, ED are greater than C D.
To each of these equals add D B.
(CE, ED, and DB make CE and E B.)
(ƒ) .. BE and EC are greater than BD
and DC (e).
Now, comparing results (d) and (ƒ) we have
.. BA, AC are greater than BD, DC.
Proof (of 6).
Q. E. D.
EDC is a triangle, having the side ED produced to B.
.. by Euc. I. 16,
(c) The exterior angle B D C is greater than the interior opposite angle D E C.
Again, A E B is a triangle, having the side AE produced to C.
by Euc. I. 16,
(d) The exterior angle DEC is greater
Q. E. D,
EXERCISES.-I. On the opposite side of AB construct two triangles; one having the angle at the
vertex greater than ACB; the other having the angle at the vertex smaller than A CB.
II. Given.-AC, CB together less than AD, DB.
Required. To prove that angle ACB is greater than angle ADB. (Suppose the triangle ACB placed on the other side of A B.)
THEOREM (Euclid 1. 4).
Repeat.-Axioms 8, 8a, 10.
If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases, or third sides, equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely, those to which the equal sides are opposite.
This is a very important and useful proposition, and it is necessary to thoroughly understand it. Here are the parts given and required in a tabular form.
Given. The two triangles ABC, DEF, having
(a) The two sides AB and AC equal to the two sides DE and DF, each to each, and (b) The angle B A C equal to the angle EDF.
Required. To prove that
(a) The base B C is equal to the base EF; (b) The triangle ABC is equal to the triangle DEF;
(c) The angle ABC is equal to the angle DEF; (d) The angle ACB is equal to the angle DFE.
Proof (by superposition). (The different steps of this proof can be more readily made clear by pasteboard triangles, one of which should actually be applied to the other.)
Let the triangle ABC be applied to (placed upon) the triangle DEF, so that the point A falls upon the point B; and the line AB falls upon DE. Then
(a) The point B shall coincide with the point E because AB is equal to DE (Axiom 8a). (If AB were less than DE, B would fall between D and E; if AB were greater than DE, B would fall beyond E.) Since AB falls on DE,
(b) AC will fall on DF,
because angle BAC is equal to angle EDF. (If angle B A C were less than angle EDF, AC would fall between D E and DF; if angle B A C were greater than angle EDF, AC would fall beyond DF.)
Since AC falls on DF,
(c) The point C will coincide with the point F because A C is equal to D F. The point B coincides with E (a).
The point C coincides with F (c).
(d). The line BC will coincide with EF, because two straight lines cannot enclose a space (Axiom 10).
(If B coincides with E, and C with F, and the
whole line B C did not coincide with the whole
Now: The side A B coincides with DE (a).
The base B C coincides with E F (d). (e). The triangle ABC is equal to the triangle DEF, and angle ABC coincides with angle DEF
(f).. angle ABC is equal to angle DE F,
also, angle A CD coincides with angle DFE. (g) .. angle A CD is equal to angle DFE. Collecting (d), (e), (ƒ), and (g), we have
The base B C equal to the base E F (d).
The triangle ABC equal to the triangle DEF (e). The angle ABC equal to the angle DEF (f). and The angle A CB equal to the angle DFE (g). Q. E. D.
EXERCISES. (The same method of reasoning to be employed as in the above proposition.)
I. Given.-Side AB equal to side A C, and angle BAD equal to angle CAD.