In any triangle, an exterior angle (βe) 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, βe, is greater than both interior angles α and γ.
Therefore, in triangle ABC, the exterior angle βe 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 βe from now on.
We need to demonstrate that the exterior angle βe 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 BMC are congruent because they have two congruent sides BM ≅ CM and AM ≅ ME, and the angle between them is congruent θ1 ≅ θ2.
$$ AMC \cong BMC $$
Thus, triangles AMC and BMC 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 βe because segment BE divides angle βe
$$ \delta < \beta_e $$
Knowing that angles γ and δ are congruent (γ ≅ δ), we can deduce that angle γ is also smaller than the exterior angle βe
$$ \gamma < \beta_e $$
Therefore, the exterior angle βe is greater than the interior angle γ
Second Part
Next, we need to perform a similar operation to show that the exterior angle βe 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 βe with angle δ, extend segment BC
Then, knowing that vertical angles are equal, project an angle congruent to βe:
It becomes evident that the exterior angle βe is greater than angle δ because segment BF divides angle βe
$$ \delta < \beta_e $$
Knowing that angle δ is congruent to interior angle α (δ ≅ α), we deduce that the exterior angle βe is greater than the interior angle α.
Conclusion
In a triangle, the exterior angle βe 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 βe
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 βe = βe' + βe''
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 βe = βe' + βe'' 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 βe 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 βe = α + β is equal to the sum of the non-adjacent interior angles of the triangle.
And so on.
