MathHub US Geometry Chapter 4

Chapter 4: Triangle Relationships

High School Geometry · Unit 3: Triangles · 3 interactive diagrams · 8 practice problems

Learning Objectives

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 $\angle C$.

$\angle C = 180° - 47° - 68° = \mathbf{65°}$

TRY IT

A triangle has angles in the ratio $2:3:4$. Find all three angles.

Show Answer
The angles sum to $180°$: $2k + 3k + 4k = 180° \Rightarrow 9k = 180° \Rightarrow k = 20°$. The angles are $\mathbf{40°, 60°, 80°}$.

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: $52° + 63° + (180° - 115°) = 52° + 63° + 65° = 180°$ ✓

TRY IT

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.

Show Answer
The exterior angle at $R$ equals the sum of the two remote interior angles: $38° + 74° = \mathbf{112°}$. Also, $\angle R = 180° - 38° - 74° = 68°$, so the exterior angle = $180° - 68° = 112°$ ✓

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.

TRY IT

Two sides of a triangle are 7 and 13. What values are possible for the third side $x$?

Show Answer
We need $7 + 13 > x$ and $7 + x > 13$ and $13 + x > 7$. From the first: $x < 20$. From the second: $x > 6$. From the third: always true. So $\mathbf{6 < x < 20}$.

Side–Angle Relationship

Within a triangle, angles and opposite sides are related in a predictable order:

Side–Angle Inequality

In any triangle:

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:

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).

TRY IT

$\overline{DE}$ is a midsegment of $\triangle ABC$ with $DE = 5x - 3$ and $BC = 7x + 4$. Find $x$ and $DE$.

Show Answer
By the Midsegment Theorem: $DE = \tfrac{1}{2}BC$, so $5x - 3 = \tfrac{1}{2}(7x + 4) \Rightarrow 10x - 6 = 7x + 4 \Rightarrow 3x = 10 \Rightarrow x = \tfrac{10}{3}$. Then $DE = 5(\tfrac{10}{3}) - 3 = \tfrac{50}{3} - 3 = \tfrac{41}{3} \approx 13.7$.

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

1

In $\triangle DEF$, $\angle D = 3x + 5$, $\angle E = 2x - 7$, $\angle F = x + 20$. Find the measure of each angle.

Show Solution
$(3x+5) + (2x-7) + (x+20) = 180 \Rightarrow 6x + 18 = 180 \Rightarrow x = 27$. So $\angle D = 86°$, $\angle E = 47°$, $\angle F = 47°$.
2

An exterior angle of a triangle is $4x + 10$. The two remote interior angles are $x + 30$ and $2x - 5$. Find $x$.

Show Solution
$4x + 10 = (x+30) + (2x-5) \Rightarrow 4x + 10 = 3x + 25 \Rightarrow x = 15$. Check: exterior = $70°$, remote angles = $45°$ and $25°$; $45+25=70$ ✓
3

Can a triangle have sides 4, 11, and 7? Explain. What about 5, 12, and 13?

Show Solution
4, 11, 7: $4 + 7 = 11$, not $> 11$. Not a valid triangle (degenerate). 5, 12, 13: $5 + 12 = 17 > 13$ ✓ — valid triangle (it's actually a right triangle since $5^2 + 12^2 = 169 = 13^2$).
4

In $\triangle XYZ$, $XY = 5$, $YZ = 9$, $XZ = 7$. Order the angles from smallest to largest.

Show Solution
Largest side ($YZ = 9$) is opposite the largest angle ($\angle X$). Middle ($XZ = 7$) is opposite middle ($\angle Y$). Smallest ($XY = 5$) is opposite smallest ($\angle Z$). Order: $\mathbf{\angle Z < \angle Y < \angle X}$.
5

Two sides of a triangle are 11 and 20. Find all integers that could be the length of the third side.

Show Solution
Need $|20 - 11| < x < 20 + 11$, so $9 < x < 31$. The integer values are $\mathbf{10, 11, 12, \ldots, 30}$ (21 values).
6

$\overline{MN}$ is a midsegment of $\triangle ABC$. If $MN = 3x + 1$ and $BC = 5x + 8$, find $MN$.

Show Solution
$3x + 1 = \tfrac{1}{2}(5x + 8) \Rightarrow 6x + 2 = 5x + 8 \Rightarrow x = 6$. Then $MN = 3(6) + 1 = \mathbf{19}$. Check: $BC = 5(6)+8 = 38 = 2 \times 19$ ✓
7

Proof: Using the Angle Sum Theorem, prove that each angle of an equilateral triangle measures 60°.

Show Solution
Given: $\triangle ABC$ is equilateral, so $AB = BC = CA$. Prove: $\angle A = \angle B = \angle C = 60°$. Since all sides are equal, all opposite angles are equal (isosceles theorem applied to each pair): $\angle A = \angle B = \angle C$. By the Angle Sum Theorem: $\angle A + \angle B + \angle C = 180°$, so $3\angle A = 180°$, giving $\angle A = \angle B = \angle C = 60°$. $\square$
8

$\triangle ABC$ 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 $\angle BDE$.

Show Solution
By the Midsegment Theorem, $DE = \tfrac{1}{2} \cdot AC = 12$. Since $\overline{DE} \parallel \overline{AC}$, $\angle BDE$ corresponds to $\angle BAC$ (corresponding angles with transversal $\overline{AB}$). Wait — $\angle DAC = \angle BAC = 35°$, so $\angle BDE = 35°$ (corresponding angles: $\overline{DE} \parallel \overline{AC}$).

📋 Chapter Summary

Triangle Centers

Circumcenter

Intersection of the three perpendicular bisectors. Equidistant from all three vertices. Center of the circumscribed circle.

Incenter

Intersection of the three angle bisectors. Equidistant from all three sides. Center of the inscribed circle.

Centroid

Intersection of the three medians. Located $\frac{2}{3}$ of the way from each vertex to the opposite midpoint. Center of gravity.

Orthocenter

Intersection of the three altitudes. Location varies: inside (acute), on hypotenuse (right), outside (obtuse).

Inequalities in Triangles

Triangle Inequality

The sum of any two sides must be greater than the third side: $a + b > c$, $a + c > b$, $b + c > a$.

Midsegment Theorem

A midsegment connecting midpoints of two sides is parallel to the third side and half its length.

📘 Key Terms

MedianA segment from a vertex to the midpoint of the opposite side. Three medians meet at the centroid.
AltitudeA perpendicular segment from a vertex to the opposite side (or its extension). Three altitudes meet at the orthocenter.
MidsegmentA segment connecting the midpoints of two sides of a triangle. Parallel to and half the length of the third side.
CircumcenterThe point equidistant from all three vertices of a triangle. Center of the circumscribed circle.
IncenterThe point equidistant from all three sides of a triangle. Center of the inscribed circle.
Triangle InequalityStates that the sum of any two sides of a triangle must be greater than the third side.
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