3.4 Angles of a Triangle
By definition, a triangle is a figure formed by the union of three segments whose endpoints are three noncollinear points. Each triangle has three interior angles. If two of the sides of the triangle are congruent, then the triangle is an isosceles triangle. If all three sides are congruent, then the triangle is an equilateral triangle.
- Theorem 3.10: The
sum of the measures of the interior angles of a triangle is 180.
Given: $\triangle ABC$ Prove: $m\angle A+m\angle B+m\angle C=180$ |
Proof: Given $\triangle ABC$. By the Parallel Postulate, through $B$ we can introduce a line $\ell$ that is parallel to $\overline{AC}$. $\angle 1$, $\angle 2$, and $\angle 3$ form a straight angle, so $m\angle 1+m\angle 2+m\angle 3=180$. Since $\ell$ is parallel to $\overline{AC}$, $\angle 1\cong\angle 4$ and $\angle 3\cong\angle 5$ (they are alternate interior angles). By substitution, we can now say that $m\angle 4+m\angle 3+m\angle 5=180$, thus proving that the sum of the measures of the angles of a triangle is equal to 180.
There are two immediate consequences of this theorem. The first has to do with an exterior angle of a triangle; the second has to do with the angles of two triangles. The proofs of both theorems will be left for the exercises.
- Theorem 3.11: The Exterior
Angle Theorem
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles of the triangle.
Restatement: In the figure, $m\angle 1=m\angle B+m\angle C$ |
- Theorem 3.12: The
No-Choice Theorem
If two angles of one triangle are congruent, respectively, to two angles of a second triangle, then the third angles are congruent.
There is now an interesting theorem that we can prove about three lines that are parallel.
- Theorem 3.13:
If three parallel lines cut off congruent segments on one transversal,
then they cut off congruent segments on every transversal.
Given: $\ell$, $m$, and $n$
are parallel lines and $\overline{AB}\cong\overline{BC}$. Prove: $\overline{PQ}\cong\overline{QR}$ |
Proof: We are given that lines $\ell$, $m$, and $n$ are parallel and that $\overline{AB}\cong\overline{BC}$. By the Parallel Postulate, we introduce the line through $Q$ that is parallel to $\overline{ABC}$, as shown below. This line will intersect lines $\ell$, $m$, and $n$ in points $X$, $Q$, and $Y$. $AXQB$ and $BQYC$ are now quadrilaterals in which both pairs of opposite sides are parallel. In the last section, we proved in an example that when a quadrilateral has both pairs of opposite sides parallel, then both pairs of opposite sides would also be congruent. Hence we know that $\overline{XQ}\cong\overline{AB}$ and $\overline{QY}\cong\overline{BC}$.
There is a very important special case of Theorem 3.12. If point $P$ coincides with point $A$, then the parallel line through $A$ and $P$ might not be drawn in the diagram (but it exists, nonetheless). This special case isn’t really a new theorem, but it is a result worth identifying. Such a result — an easy consequence of the main theorem, is often called a corollary. Thus we have the following corollary to Theorem 3.12.
- Corollary:
If a line through the midpoint of one side of a triangle is parallel to
a second side of the triangle, then it passes through the midpoint of
the third side.