Prove that in any triangle, the outer angle at a vertex is equal to the sum of the inner angles at the other two vertices.

Let us consider Δ ABC and ∠ ACD is and exterior angle

To show: ∠ ACD = ∠ A + ∠ B

Through C draw CE parallel to AB

Proof:

∠ A = ∠ y (AB || CE and AD is a transversal)

∠ B = ∠ x (AB || CE and AC is a transversal; alternate angles are equal)

∠ 1 + ∠ 2 = ∠ x + ∠ y

Now, ∠ x + ∠ y = ∠ ACD

Hence, ∠ 1 + ∠ 2 = ∠ ACD

∠ ACD = ∠ A + ∠ B

Hence Proved.