Exterior Angle Theorem for Triangles

In any triangle, an exterior angle (β_e) is greater than either of the non-adjacent interior angles (α and γ)

because every exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles $$ \beta_e \cong \alpha + \gamma $$

This is a universal theorem that applies to any triangle.

For example, in triangle ABC, the exterior angle at vertex B, β_e, is greater than both interior angles α and γ.

Therefore, in triangle ABC, the exterior angle β_e is larger than any interior angle other than β.

Proof

A] The Exterior Angle is Greater Than the Non-Adjacent Interior Angles

Consider a generic triangle ABC

Let's examine the exterior angle at vertex B, which we'll call β_e from now on.

We need to demonstrate that the exterior angle β_e is greater than the non-adjacent interior angles α and γ.

First Part

Identify the midpoint M of side BC adjacent to angle β and draw the median AM.

This gives us two congruent segments, BM and CM

$$ BM \cong CM $$

Extend segment AM by adding another segment ME such that AM = ME, effectively doubling its length.

Recapping, the following sides are congruent:

$$ BM \cong CM $$

$$ AM \cong ME $$

Now, connect points B and E by adding segment BE to the construction.

At point M, two segments intersect forming two congruent vertical angles θ₁ ≅ θ₂

$$ \theta_1 \cong \theta_2 $$

Therefore, according to the first congruence theorem, triangles AMC and BME are congruent because they have two congruent sides BM ≅ CM and AM ≅ ME, and the angle between them is congruent θ₁ ≅ θ₂.

$$ AMC \cong BME $$

Thus, triangles AMC and BME have congruent sides and angles in the same order.

To conclude the proof, it's important to note that angles γ and δ are congruent:

$$ \gamma \cong \delta $$

Angle δ is smaller than the exterior angle β_e because segment BE divides angle β_e

$$ \delta < \beta_e $$

Knowing that angles γ and δ are congruent (γ ≅ δ), we can deduce that angle γ is also smaller than the exterior angle β_e

$$ \gamma < \beta_e $$

Therefore, the exterior angle β_e is greater than the interior angle γ

Second Part

Next, we need to perform a similar operation to show that the exterior angle β_e is also greater than the other non-adjacent interior angle, α.

Identify the midpoint M of segment AB adjacent to angle β and draw the median CM.

Consequently, the two segments AM and BM are congruent:

$$ \overline{AM} \cong \overline{BM} $$

Extend segment CM by adding another segment MF of equal length:

Therefore, segments CM and MF are congruent:

$$ \overline{CM} \cong \overline{MF} $$

Recapping, the following sides are congruent:

$$ \overline{AM} \cong \overline{BM} $$

$$ \overline{CM} \cong \overline{MF} $$

Connect points B and F by adding segment BF to the construction.

At point M, two segments AB and CF intersect forming two congruent vertical angles θ₁ ≅ θ₂

$$ \theta_1 \cong \theta_2 $$

Thus, according to the first congruence theorem, triangles AMC and BMF are congruent because they have two congruent sides CM ≅ FM and AM ≅ BM, and the angle between them is congruent θ₁ ≅ θ₂.

$$ AMC \cong BMF $$

Thus, triangles AMC and BMF have congruent sides and angles in the same order.

Importantly , angle α is congruent to angle δ (purple):

$$ \alpha \cong \delta $$

To compare angle β_e with angle δ, extend segment BC

Then, knowing that vertical angles are equal, project an angle congruent to β_e:

It becomes evident that the exterior angle β_e is greater than angle δ because segment BF divides angle β_e

$$ \delta < \beta_e $$

Knowing that angle δ is congruent to interior angle α (δ ≅ α), we deduce that the exterior angle β_e is greater than the interior angle α.

Conclusion

In a triangle, the exterior angle β_e is greater than the non-adjacent interior angles α and γ.

This proves the exterior angle theorem for a triangle.

B] The Exterior Angle is Equal to the Sum of the Non-Adjacent Interior Angles

Next, we need to show that the exterior angle is equal to the sum of the non-adjacent interior angles.

Consider triangle ABC and the exterior angle β_e

Draw lines passing through the segments of the triangle and name them a, b, c:

Draw line d parallel to segment AC passing through vertex B of the triangle.

Now, we can consider the exterior angle as the sum of two angles β_e = β_e' + β_e''

Angle β'_e is an alternate interior angle of the parallel lines a || d cut by line c.

By the parallel lines theorem, the alternate interior angles γ and β'_e are congruent, i.e., γ ≅ β'_e.

Angle β''_e is a corresponding angle of the parallel lines a || d cut by line b.

By the parallel lines theorem, the corresponding angles α and β''_e are congruent, i.e., α ≅ β''_e.

Knowing that the exterior angle is the sum β_e = β_e' + β_e'' and that angles γ ≅ β'_e and α ≅ β''_e are congruent:

$$ \begin{cases} \beta_e \cong \beta'_e + \beta''_e \\ \\ \gamma \cong \beta'_e \\ \\ \alpha \cong \beta''_e \end{cases} $$

We deduce that the exterior angle β_e is equal to the sum of the non-adjacent interior angles α + β of the triangle.

$$ \beta_e \cong \beta'_e + \beta''_e $$

$$ \beta_e \cong \beta + \alpha $$

Therefore, the exterior angle β_e = α + β is equal to the sum of the non-adjacent interior angles of the triangle.

Notes

Additional observations and corollaries related to the topic.

• The sum of two interior angles is always less than a straight angle
  This fundamental corollary follows from the fact that an exterior angle of a triangle is always greater than either of the two non-adjacent interior angles. In other words, $ \alpha + \gamma < 180° $.

  

  Proof. By the exterior angle theorem, we know that the exterior angle \( \beta_e \) is greater than the non-adjacent interior angle \( \gamma \): $$ \gamma < \beta_e $$ Adding \( \beta \) to both sides of the inequality gives: $$ \gamma + \beta < \beta_e + \beta $$ Since \( \beta \) and \( \beta_e \) are supplementary, their sum forms a straight angle: $$ \beta + \beta_e = 180^\circ $$ Substituting this into the inequality, we get: $$ \gamma + \beta < 180^\circ $$ This confirms that the sum of any two interior angles in a triangle is always less than a straight angle. Using the same argument, we can also conclude that the other two pairs, $ \alpha + \gamma $ and $ \alpha + \beta $, are also less than $ 180^\circ $.

• A triangle can have at most one right or one obtuse angle
  This follows directly from the fundamental property that the sum of a triangle’s interior angles is always 180°.

  Note. If a triangle had two right angles (90° + 90° = 180°), there would be no room for a third angle, which is impossible. Likewise, if a triangle had two obtuse angles (each greater than 90°), their sum would exceed 180°, which contradicts the angle sum property. Therefore, a triangle can have at most one right angle or one obtuse angle. The remaining two angles must be acute (less than 90°) to ensure the total remains 180°.

And so on.