122. Cor. 3. If 2 angles of a triangle are equal, each is an acute angle.

= 180°

64

123. Cor. 4. If 2 angles of a triangle are given, the 3d angle may easily be found, for it is equal to 2 R. A. less the sum of the given angles. If one angle=64°, and another = 70°, then the 3d angle 70° 46°. Thus if two angles of one triangle are equal to two angles of another, then the third angle of the one must be equal to the third angle of the other.

=

124. The exterior angle of a triangle, that is, the angre made by producing one of the sides of a triangle, is equal to the sum of the two opposite interior angles, (fig. 88.) Angle ACD=angle ABC + angle CAB.

Demi. ACD + ACB=2 R. A.; ABC+CAB+ ACB 2 R. A.; therefore ACD+ACB= ABC + CAB ACB. Subtract ACB from each side, and we have ACD ABC+ CAB.

=

125. The three exterior angles of a triangle are together equal to 4 R. A. For the sum of each pair of adjacent angles is 2 R. A.; in all 3 × 26 R. A. The interior angles together-2 R. A., therefore the exterior angles together 6-24 R. A.

=

3. OF EQUAL TRIANGLES.

126. Two triangles are equal, if 2 sides and the included angle of the one are equal to 2 sides and the included angle of the other, each to each.

Dem. Let BC= EF, BA= ED, and angle BE, (fig. 62.) Now suppose the triangle ABC to be placed upon DEF, so that point B shall fall upon E, and C

upon F; this is possible, because BC= EF; in this case BC will coincide with EF. Because angle B=E, BA must take the direction ED, and because BA=ED, the point A must fall on D. Thus the end-points C and A of the line CA fall upon the points F and D, which are the ends of the straight line FD, consequently CA and FD must coincide. Thus the two triangles coincide throughout, and consequently they are equal.

127. Two triangles are equal if two angles and the interjacent (lying between) side of the one are equal to two angles and the interjacent side of the other, each to each.

Dem. (fig. 62.) Let angle BE, angle C=F, and side BC=EF. Suppose the triangle ABC to be laid upon DEF, so that BC shall coincide with EF. Because angle B=E, BA must be in the direction of ED, and the point A must fall in ED; because angle C F, CA must be in the direction of FD, and point A must fall in FD. Thus the point A is in both the lines FD and ED; it can only be at their point of intersection D. Consequently the two triangles coincide throughout, and are equal.

=

128. Remark. In both these cases of equal triangles, equal angles are opposite to equal sides.

129. In every isosceles triangle the angles opposite to the equal sides are equal, (fig. 86.)

If ACB is an isosceles triangle of which the sides AC and CB are equal, then the angles A and B are likewise equal.

Dem. Bisect the third angle C by a straight line. which shall meet the opposite side at the point D. In

the triangles ACD and BCD, the sides AC and CB are equal by supposition, CD is common, and the angle ACDBCD by construction. Therefore triangle ACD=BCD, (126,) consequently angles A and B are equal.

130. Cor. 1. If the two equal sides of an isosceles triangle be produced, the angles formed without the triangle will be equal, for each of them, together with one of the equal angles A and B, is equal to 2 R. A. (104.)

131. Cor. 2. Equilateral triangles are also equiangular; and each angle is equal to 60°.

132. Cor. 3. If one angle of an isosceles triangle be given, the others may easily be found. Let one of the angles at the base be given; because the other angle at the base is equal to this, the sum of the two subtracted from 180° will give the third angle. If the angle opposite to the base is given, subtract it from 180°, divide the remainder by 2, and the quotient will be the size of each angle at the base.

The base of an isosceles triangle is the side which is not equal to one of the others. In triangles not isosceles either side may be taken for the base. The vertex of the angle opposite to the base is sometimes called the vertex of the triangle.

133. From the equality of the triangles ACD and BCD the following propositions may be deduced.

1. The straight line which bisects the angle C is perpendicular to the base AB. Because the adjacent angles CDA and CDB are equal and therefore right angles.

2. A straight line drawn from the vertex of an isos

celes triangle perpendicular to the base, bisects that base.

3. If a perpendicular erected upon the middle point of one side of a triangle meets the vertex of the opposite angle, then the other two sides are equal to each other.

4. A perpendicular erected upon the middle point of the base of an isosceles triangle, bisects the opposite angle.

134. If two angles of a triangle are equal, the sides opposite to these angles are also equal.

Dem. (fig. 86.) Let angle A=B. Upon the middle point of the base AB erect a perpendicular to AB. If this perpendicular passes through the vertex of the triangle, then AC= BC, (133. 3.) Suppose it does not pass through the vertex, but meets the side AC at E. Draw EB. Triangles EAD and EBD are equal, (126 ;) consequently angle EAD EBD. But angle EAD= CBD; consequently angle EBD CBD; that is, a part is equal to the whole, which is absurd. Therefore, the perpendicular CD does pass through the vertex of the triangle, and the sides AC and BC are equal.

135. Two right angled triangles are equal, if two sides of the one are equal to two sides of the other, each to each.

Dem. If the equal sides include the right angles, then the equality of the triangles has already been demonstrated, (126.)

But let us suppose the hypothenuse AB and side BC of triangle ABC (fig. 89. 2.) to be respectively equal to the hypothenuse DE and side EF of triangle DEF. Suppose the triangle ABC to be placed so that BC shall coincide with EF. AB will take the direction of EG, and, because the angles DFE and ACB are right angles,

AC will coincide with DF produced; suppose with FG. We shall thus have an isosceles triangle DEG. Now EF is perpendicular to DG, and therefore bisects it, (133. 2;) thus GF=AC=DF. Therefore the triangles ABC and DEF are equal, (126.)

136. We are now prepared to demonstrate another principle, viz., that any point in a perpendicular EC (fig. 67,) erected at the middle of a straight line AB, is at equal distances from the two extremities of this line. Let us take the point D, and draw the lines DA and DB. The triangles DAC and DAB are equal; for AC and CB are equal by supposition, and CD is common to both. Therefore the sides AD and DB are equal, which was to be demonstrated.

137. Cor. Two oblique lines, as AD and DB, drawn at equal distances from a perpendicular, are equal.

138. Two triangles are equal if three sides of the one are equal to three sides of the other, each to each, (fig. 89. 1.) If AB-DE, ACDF, and BC= EF, then is triangle ABCDEF.

Dem. Suppose the triangle ABC to be moved from its place so that BC shall coincide with EF, and the triangle ABC shall occupy the space GEF; BA coinciding with EG, and CA with FG. Draw DG. The angles C and F may be either acute, right, or obtuse, and each case will be considered separately.

1. Acute angles, (fig. 89. 1.) By supposition ED= EG and FD=FG, therefore angle EDG=EGD, and angle FDG=FGD, (129 ;) consequently angle EDG +FDG=EDF= angle EGD+DGF= EGF. Now