Chapter 4: Triangle Relationships
Learning Objectives
- Apply the Triangle Angle Sum Theorem to find missing angles
- Use the Exterior Angle Theorem and its corollary
- Apply the Triangle Inequality Theorem to determine valid triangles
- Relate sides and angles: the longest side is opposite the largest angle
- Apply the Triangle Midsegment Theorem
- Write geometric proofs using these theorems
4.1 Triangle Angle Sum Theorem
The most fundamental fact about triangles is that their interior angles always add up to exactly 180°. This is true for every triangle, regardless of its shape or size.
Theorem 4.1 — Triangle Angle Sum Theorem
The sum of the interior angles of any triangle equals $180°$:
$$\angle A + \angle B + \angle C = 180°$$
Proof sketch: Draw a line through vertex $C$ parallel to $\overline{AB}$. The three angles at $C$ form a straight angle (180°) by the alternate interior angles theorem and substitution.
Example 4.1 — Finding a Missing Angle
In $\triangle ABC$, $\angle A = 47°$ and $\angle B = 68°$. Find $\angle C$.
$\angle C = 180° - 47° - 68° = \mathbf{65°}$
A triangle has angles in the ratio $2:3:4$. Find all three angles.
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Drag the vertices of the triangle to change its shape — the angle sum always stays 180°. Angles are labeled $\alpha$, $\beta$, $\gamma$.
Figure 4.1 — Triangle Angle Sum = 180°
4.2 Exterior Angle Theorem
An exterior angle of a triangle is formed when one side is extended beyond a vertex. It is supplementary to the adjacent interior angle.
Theorem 4.2 — Exterior Angle Theorem
The measure of an exterior angle of a triangle equals the sum of the two non-adjacent (remote) interior angles:
$$\angle 1 = \angle A + \angle B$$
where $\angle 1$ is the exterior angle at vertex $C$, and $\angle A$, $\angle B$ are the two remote interior angles.
Example 4.2 — Using the Exterior Angle Theorem
An exterior angle of a triangle measures $115°$. One remote interior angle measures $52°$. Find the other remote interior angle.
$115° = 52° + x \Rightarrow x = \mathbf{63°}$
Check: $52° + 63° + (180° - 115°) = 52° + 63° + 65° = 180°$ ✓
In $\triangle PQR$, $\angle P = 38°$ and $\angle Q = 74°$. An exterior angle is formed at $R$ by extending side $QR$. Find the exterior angle measure.
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Proof Tip: The Exterior Angle Theorem is a corollary of the Angle Sum Theorem. Let the exterior angle be $\angle 1$ at vertex $C$. Since $\angle C + \angle 1 = 180°$ and $\angle A + \angle B + \angle C = 180°$, we get $\angle 1 = \angle A + \angle B$ by substitution.
4.3 Triangle Inequality Theorem
Not every set of three lengths can form a triangle. The Triangle Inequality gives us a precise rule for when three lengths form a valid triangle.
Theorem 4.3 — Triangle Inequality Theorem
The sum of any two side lengths of a triangle must be greater than the third side:
$$a + b > c, \quad a + c > b, \quad b + c > a$$
Practical test: Check only that the sum of the two shorter sides exceeds the longest side.
Example 4.3 — Can These Form a Triangle?
(a) Sides 5, 8, 12: Is $5 + 8 > 12$? $13 > 12$ ✓ — Yes, valid triangle.
(b) Sides 3, 4, 9: Is $3 + 4 > 9$? $7 > 9$? No — not a valid triangle.
(c) Sides 6, 10, 10: Is $6 + 10 > 10$? $16 > 10$ ✓ — Yes, valid triangle.
Two sides of a triangle are 7 and 13. What values are possible for the third side $x$?
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Side–Angle Relationship
Within a triangle, angles and opposite sides are related in a predictable order:
Side–Angle Inequality
In any triangle:
- The longest side is opposite the largest angle.
- The shortest side is opposite the smallest angle.
- If two sides are equal, the opposite angles are equal (isosceles triangle theorem).
Triangle with labeled sides and angles — observe how the longest side is always opposite the largest angle.
Figure 4.2 — Side–Angle Relationship in Triangles
4.4 Triangle Midsegment Theorem
A midsegment of a triangle connects the midpoints of two sides. It has remarkable properties.
Theorem 4.4 — Triangle Midsegment Theorem
The segment connecting the midpoints of two sides of a triangle (the midsegment) is:
- Parallel to the third side
- Half the length of the third side
If $M$ and $N$ are midpoints of $\overline{AB}$ and $\overline{AC}$, then $\overline{MN} \parallel \overline{BC}$ and $MN = \tfrac{1}{2} BC$.
Example 4.4 — Using the Midsegment Theorem
In $\triangle ABC$, $M$ is the midpoint of $\overline{AB}$ and $N$ is the midpoint of $\overline{AC}$. If $BC = 28$, find $MN$.
$MN = \tfrac{1}{2} \cdot BC = \tfrac{1}{2} \cdot 28 = \mathbf{14}$
Also, $\overline{MN} \parallel \overline{BC}$, so $\angle AMN = \angle ABC$ (corresponding angles).
$\overline{DE}$ is a midsegment of $\triangle ABC$ with $DE = 5x - 3$ and $BC = 7x + 4$. Find $x$ and $DE$.
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Geometry Proof Strategy: The Midsegment Theorem can be proved using coordinate geometry — place the triangle with $A$ at the origin, find the midpoints $M$ and $N$, show $\overline{MN}$ has the same slope as $\overline{BC}$ (parallel), and compute $MN = \tfrac{1}{2}BC$ using the distance formula.
Interactive triangle with midsegment $\overline{MN}$ shown in blue — the midsegment is parallel to the base and half its length.
Figure 4.3 — Triangle Midsegment: Parallel and Half the Base
Practice Problems
In $\triangle DEF$, $\angle D = 3x + 5$, $\angle E = 2x - 7$, $\angle F = x + 20$. Find the measure of each angle.
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An exterior angle of a triangle is $4x + 10$. The two remote interior angles are $x + 30$ and $2x - 5$. Find $x$.
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Can a triangle have sides 4, 11, and 7? Explain. What about 5, 12, and 13?
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In $\triangle XYZ$, $XY = 5$, $YZ = 9$, $XZ = 7$. Order the angles from smallest to largest.
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Two sides of a triangle are 11 and 20. Find all integers that could be the length of the third side.
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$\overline{MN}$ is a midsegment of $\triangle ABC$. If $MN = 3x + 1$ and $BC = 5x + 8$, find $MN$.
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Proof: Using the Angle Sum Theorem, prove that each angle of an equilateral triangle measures 60°.
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$\triangle ABC$ has midsegment $\overline{DE}$ where $D$ is the midpoint of $\overline{AB}$ and $E$ is the midpoint of $\overline{BC}$. If $AC = 24$ and $\angle DAC = 35°$, find the length $DE$ and the measure of $\angle BDE$.
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📋 Chapter Summary
Triangle Centers
Intersection of the three perpendicular bisectors. Equidistant from all three vertices. Center of the circumscribed circle.
Intersection of the three angle bisectors. Equidistant from all three sides. Center of the inscribed circle.
Intersection of the three medians. Located $\frac{2}{3}$ of the way from each vertex to the opposite midpoint. Center of gravity.
Intersection of the three altitudes. Location varies: inside (acute), on hypotenuse (right), outside (obtuse).
Inequalities in Triangles
The sum of any two sides must be greater than the third side: $a + b > c$, $a + c > b$, $b + c > a$.
A midsegment connecting midpoints of two sides is parallel to the third side and half its length.