PROOF. a= c,

b/a c/d

b = xd, 1 = 43, 42 = 4 4. (108. If two lines intersect, the vertical angles are equal.)

Moreover, since, by hypothesis, a = 1, their supplements are equal, or b = 2.

(98. All straight angles are equal.)

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

4 ¥4

= b + 43.

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

SECOND CASE. If, instead of two corresponding, we have given two alternate angles equal, we substitute for one its vertical, which gives the First Case again.

THIRD CASE.

HYPOTHESIS. a + 4 4 = st. 4.

But 4 + 1 = st. ¥,

.. 4 a = 41,

which gives again the First Case.

FOURTH CASE. HYPOTHESIS. 1 + 4 d = st. 4.

But a + d = st. 4,

III. Triangles.

114. An Equilateral Triangle is one in which the three sides are equal.

A

A

115. An Isosceles Triangle is one which has two sides equal.

116. A Scalene Triangle has no two

sides equal.

117. When one side of a triangle has to be distinguished from the other two, it may be called the Base; then that one of the vertices opposite the base is called the Vertex.

118. When we speak of the angles of a triangle, we mean the three interior angles.

119. A Right-angled Triangle has one of its angles a right angle. The side opposite the right angle is called the Hypothe

nuse.

120. An Obtuse-angled Triangle has one of its angles obtuse. 121. An Acute-angled Triangle has all three angles acute.

122. An Equiangular Triangle is one which has all three angles equal.

123. When two triangles have three angles of the one. equal respectively to the three angles of the other, a pair of equal angles are called Homologous Angles. The pair of sides opposite homologous angles are called Homologous Sides.

THEOREM VI.

124. Two triangles are congruent if two sides and the included angle in the one are equal respectively to two sides and the included angle in the other.

HYPOTHESIS. ABC and LMN two triangles, with AB = LM, BC MN, B = 4 M.

CONCLUSION. The two triangles are congruent; or, using ▲ for "triangle," and for "congruent,"

A ABC = A LMN.

PROOF. Apply the triangle LMN to ▲ ABC in such a manner that the vertex M shall rest on B, and the side ML on BA, and the point N on the same side of BA as C.

=

Then, because the side ML equals BA, the point I will rest upon A; because M B, the side MN will rest upon the line BC; because the side MN equals BC, the point N will rest upon the point C. Now, since the point L rests upon A, and the point N rests upon C, therefore the side LN coincides with the side AC.

(95. If two lines have two points in common, the two sects between those points

coincide.)

Therefore every part of one triangle will coincide with the corresponding part of the other, and the two are congruent.

(94. Magnitudes which will coincide are congruent.)

125. In any pair of congruent triangles, the homologous sides are equal.

THEOREM VII.

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

HYPOTHESIS. ABC a triangle, with AB = BC.
CONCLUSION. ¥ A =

XC.

A

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',

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

(87. Things equal to the same thing are equal to one another.)

127. COROLLARY. Every equilateral triangle is also equiangular.

EXERCISES. 2. The bisectors of vertical angles are in the 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.

HYPOTHESIS. ABC and LMN two triangles, with A = L, C = N, AC = LN.

CONCLUSION. The two triangles are congruent.

A ABC & 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.

.. side NM will rest upon the line CB.

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

..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.