3. Show that if the perpendicular from the vertex to the base of a triangle bisects the base, the triangle is isosceles.

4. How many degrees are there in the exterior angle at each vertex of an equiangular triangle ?

5. Show that the bisectors of the equal B angles of an isosceles triangle form with the base another isosceles triangle; that is, DBC is isosceles.

6. What are the relative values of the vertical angles D and A in the above?

7. Show that a straight line parallel to the base of an isosceles triangle makes equal angles with its sides, or D=

E.

A

D

D

E

PROPOSITION XVIII. THEOREM.

95. Of two sides of a triangle, that is greater which is opposite the greater angle.

In the triangle ABC let angle ACB be greater than angle B. To prove that AB> AC.

From C draw CE, making ≤ ECB = ≤ B.

Then the triangle BEC is isosceles and the side EB EC. Add AE,

AE + EB

AE + EC,

.. AB, which is equal to AE + EC, is greater than AC.

Q.E.D.

EXERCISE.

1. Show that of two angles of a triangle, that is the greater which is opposite the greater side.

PROPOSITION XIX. THEOREM.

96. The three bisectors of the three angles of a triangle meet in a point.

B

Let AO and CO be the bisectors of the angles A and C of the triangle ABC.

To prove that the bisectors meet in a point.

Suppose 40 and CO meet at O, and join BO.

Let fall the perpendiculars, OP, OH, and OK, forming the right triangles AOP, AOH, COP, and COK.

The triangles AOP and AOH are equal (by 89), having the hypotenuse A0 common, and ZOAP=OAH; therefore OP = OH.

or

For the same reason, the triangles OPC and OKC are equal, OP = OK,

The two right triangles BOH and BOK have BO common, and OH= OK; therefore (by 88) they are equal; that is,

or

/ HBO = / KBO,

BO is a bisector of B.

.. the three lines meeting in O are bisectors of the angles.

97. COR. Since OP=OK

=

OH, it is shown that the bisectors of angles are equally distant from their sides.

EXERCISES,

1. Show that the three perpendiculars erected at the middle points of the three sides of a triangle meet in a point, and this point is equally distant from the vertices.

SUG. See 53.

2. If an exterior angle is formed at each vertex of a triangle, their sum will be equal to four right angles.

3. Show that the perpendiculars from the vertices of a triangle to the opposite sides meet in a point.

4. Show that every point unequally distant from the sides of an angle lies outside of the bisector of that angle.

QUADRILATERALS.

DEFINITIONS.

98. A Quadrilateral is a plane figure bounded by four straight lines; as ABCD.

99. The bounding lines are called the Sides of the quadrilateral, and their points of intersection are called its Vertices

A

B

100. The Angles of the quadrilateral are the interior angles formed by the sides with each other.

101. A Diagonal of a quadrilateral is a straight line joining two vertices not adjacent; as AC.

102. Quadrilaterals are divided into classes as follows:

1st. The Trapezium (A), which has no two

of its sides parallel.

2d. The Trapezoid (B), which has two sides. parallel. The parallel sides are called the Bases, and the perpendicular distance between them the Altitude of the trapezoid.

3d. The Parallelogram (C), which is bounded by two pairs of parallel sides.

103. The side upon which a parallelogram

is supposed to stand and the opposite side are

B

C

A

called its lower and upper Bases. The perpendicular distance between the bases is the Altitude.

104. Parallelograms are divided into species as follows:

The Rhomboid (A), whose adjacent sides are not equal and whose angles are not right angles.

The Rhombus (B), whose sides are all equal. The Rectangle (C), whose angles are all right angles.

A

B

C

The Square (D), whose sides are all equal and whose angles are all equal.

105. The square is at once a rhombus and a rectangle.

D

PROPOSITION XX. THEOREM.

106. In a parallelogram the opposite sides are equal, and the opposite angles are equal.

B

с

Α

E

Let the figure ABCE be a parallelogram.

=

To prove that AB CE, and BC= AE, and B=ZE, LA=LC.

Draw the diagonal AC.

Since AB and CE are parallel and AC cuts them,

Since AE and BC are parallel and AC cuts them,

Then the triangles ABC and ACE have the side AC common, and the two adjacent angles equal; they are, therefore (by 88), equal in all their parts,

107. COR. 1. A diagonal of a parallelogram divides it into two equal triangles.

108. COR. 2. Parallel lines included between parallels are equal.