Triangle Theorems: Angle Sum, Isosceles, Midsegment and Medians

The angle sum is 180°

Given: $\triangle ABC$. Prove: $m\angle A + m\angle B + m\angle C = 180°$.

Draw line $\ell$ through $C$ parallel to $\overline{AB}$. Through a point not on a line there is exactly one parallel line — the parallel postulate — so $\ell$ exists and is unique.

Three angles now sit side by side along $\ell$ at $C$: $\angle 1$ between $\ell$ and $\overline{CA}$, then $\angle 2 = \angle ACB$ inside the triangle, then $\angle 3$ between $\overline{CB}$ and $\ell$.

Statement	Reason
$\ell \parallel \overline{AB}$	parallel postulate
$m\angle 1 + m\angle 2 + m\angle 3 = 180°$	the three angles form a straight angle
$\angle 1 \cong \angle A$	alternate interior angles, with $\overline{AC}$ as transversal
$\angle 3 \cong \angle B$	alternate interior angles, with $\overline{BC}$ as transversal
$m\angle A + m\angle ACB + m\angle B = 180°$	substitution

The whole proof is one idea: carry the two base angles up to the top vertex, where the three together make a straight line. The alternate interior angles theorem does the carrying.

Isosceles base angles are congruent

Given: $\triangle ABC$ with $AB = AC$. Prove: $\angle B \cong \angle C$.

Let $M$ be the midpoint of $\overline{BC}$ and draw $\overline{AM}$.

Statement	Reason
$AB = AC$	given
$BM = CM$	definition of midpoint
$AM = AM$	reflexive property
$\triangle ABM \cong \triangle ACM$	SSS
$\angle B \cong \angle C$	CPCTC

The added segment is what makes the proof possible. Splitting the triangle down its line of symmetry produces two triangles that can be compared, and the base angles become corresponding parts.

The converse is also true: if two angles of a triangle are congruent, the sides opposite them are congruent. Its proof uses AAS on the same two halves, with the segment drawn as the bisector of the top angle.

The exterior angle theorem

An exterior angle is formed by one side of a triangle and the extension of an adjacent side.

$$m\angle_{\text{exterior}} = m\angle A + m\angle B$$

where $A$ and $B$ are the two interior angles not next to it.

The proof is two lines. The exterior angle and $\angle C$ form a linear pair, so they sum to $180°$. The angle sum says $\angle A + \angle B + \angle C = 180°$ too. Subtracting $\angle C$ from both leaves the exterior angle equal to $\angle A + \angle B$.

Why coordinates suit the next two proofs

The midsegment theorem is about parallel segments and half a length. Both are things coordinates measure directly: parallel means equal slopes, and length comes from the distance formula.

The placement of the triangle is a choice, and a good one saves a page of algebra. Any triangle can be moved by a rigid motion so that one vertex is at the origin and one side lies along the $x$-axis — and rigid motions change no lengths or angles, so nothing proved about the placed triangle is lost.

Using $2a$, $2b$ and $2c$ for the coordinates is the second choice. Midpoints halve coordinates, and halving an even expression leaves no fractions.

The midsegment theorem

A midsegment joins the midpoints of two sides of a triangle.

Prove: it is parallel to the third side and half as long.

Place $A(0, 0)$, $B(2a, 0)$ and $C(2b, 2c)$.

Midpoints of $\overline{AC}$ and $\overline{BC}$:

$$M = \left(\frac{0 + 2b}{2}, \frac{0 + 2c}{2}\right) = (b, c) \qquad N = \left(\frac{2a + 2b}{2}, \frac{0 + 2c}{2}\right) = (a + b, c)$$

Parallel. $M$ and $N$ share the $y$-coordinate $c$, so $\overline{MN}$ is horizontal. $\overline{AB}$ lies on the $x$-axis, also horizontal. Both slopes are $0$, so the segments are parallel.

Half as long.

$$MN = (a + b) - b = a \qquad AB = 2a - 0 = 2a$$

So $MN = \tfrac{1}{2} AB$. That completes the proof — for every triangle, because $a$, $b$ and $c$ stand for any values at all.

The medians meet at one point

A median joins a vertex to the midpoint of the opposite side. A triangle has three, and they always pass through a single point, the centroid.

Take vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ and consider the point

$$G = \left(\frac{x_1 + x_2 + x_3}{3}, \; \frac{y_1 + y_2 + y_3}{3}\right)$$

The median from the first vertex ends at the midpoint of the other two, $\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right)$. Go two-thirds of the way along it:

$$x = \frac{1}{3}x_1 + \frac{2}{3}\cdot\frac{x_2 + x_3}{2} = \frac{x_1 + x_2 + x_3}{3}$$

The $y$-coordinate works the same way, so the point two-thirds along this median is exactly $G$.

Now the key observation. The formula for $G$ treats the three vertices identically — swapping their names changes nothing. So the same calculation puts $G$ two-thirds along the second median and the third. All three pass through $G$.

Worked examples

Common mistakes

Practice problems