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THEOREM VII.

126. In an isosceles triangle the angles opposite the equal sides are equal. B'

B

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HYPOTHESIS. ABC a triangle, with AB

= BC. CONCLUSION. A = XC.

PROOF. Imagine the triangle ABC to be taken up, turned over, and put down in a reversed position; and now designate the angular points A' (read A prime), B', C', to distinguish the triangle from its trace ABC left behind.

Then, in the triangles ABC, C'B'A',

AB

C'B' and BC = B'A',

since, by hypothesis, AB = CB. Also

* B = 4 B',

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* A = (124. Triangles are congruent if two sides and the included angle are equal in each.)

But *C' = XC,

... * A = *C.
(87. Things equal to the same thing are equal to one another.)

127. COROLLARY. Every equilateral triangle is also equiangular.

2. The bisectors of vertical angles are in the

EXERCISES.

same line.

THEOREM VIII.

128. Two triangles are congruent if two angles and the included side in the one are equal respectively to two angles and the included side in the other.

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HYPOTHESIS. ABC and LMN two triangles, with 4 A = 4 L, *C= XN, AC = LN.

CONCLUSION. The two triangles are congruent.

A ABC = A LMN.

PROOF. Apply the triangle LMN to the triangle ABC so that the point I shall rest upon A, and the side LN lie along the side AC, and the point M lie on the same side of AC as B. Then, because the side AC LN,

.:. point N will rest upon point C. Because L = 4 A,

side LM will rest upon the line AB.

Because N= XC,

:. side NM will rest upon the line CB.

.

Because the sides LM and NM rest respectively upon the lines AB

and CB,

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the vertex M, resting upon both the lines AB and CB, must rest

upon the vertex B, the only point common to the two lines. Therefore the triangles coincide in all their parts, and are congruent. THEOREM IX.

129. Two triangles are congruent if the three sides of the one are equal respectively to the three sides of the other.

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BC =

HYPOTHESIS. Triangles ABC and LMN, having AB = LM,

MN, and CA NL. CONCLUSION. Δ ABC LMN.

PROOF. Imagine a ABC to be applied to A LMN in such a way that LN coincides with AC, and the vertex M falls on the side of AC opposite to the side on which В falls, and join BM.

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Then in A ABM, because, by hypothesis, AB = AM,

* ABM = X AMB. (126. In an isosceles triangle the angles opposite the equal sides are equal.) And for the same reason, in A BCM, because BC CM,

X CBM = X CMB.

.

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Therefore * ABM + X CBM = X AMB + X CMB;

(88. If equals be added to equals, the sums are equal.)

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(124. Triangles are congruent if two sides and the included angle are equal in each.)

CASE II. When BM passes through one extremity of the sect AC, as C.

3

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M Then, by the first step in Case I.,

*B= XM; and for the same reason as before,

A ABC = A LMN.

CASE III. When BM falls beyond the extremity of the sect AC

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Then in A ABM, because, by hypothesis, AB = AM,

i.
* ABM =

- % AMB. And in A BCM, because BC = CM,

X CMB = X CBM. Therefore the remainders, 4 ABC = X AMC;

(89. If equals be taken from equals, the remainders are eqı

...

and so, as in Case I., A ABC = ALMN.

CHAPTER IV.

PROBLEMS.

130. A Problem is a proposition in which something is required to be done by a process of construction, which is termed the Solution

131. The solution of a problem in elementary geometry consists,

(1) In indicating how the ruler and compasses are to be used in effecting the construction required.

(2) In proving that the construction so given is correct.

(3) In discussing the limitations, which sometimes exist, within which alone the solution is possible.

PROBLEM I.

132. To describe an equilateral triangle upon a given sect.

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GIVEN, the sect AB.
REQUIRED, to describe an equilateral triangle on AB.

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