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An exterior angle of a triangle is equal to the sum of its opposite interior angles.

For vertex \(C\), the interior angle is \(∠ACB = c\) and the exterior angle is \(∠ACD = d\).

By the property, we can write that:

\(∠ACD = ∠CAB + ∠CBA\).

That is, \(d = a+b\).

Let's prove this statement through logical argument.

**An exterior angle of a triangle is equal to the sum of its interior opposite angles**.

Given:

Consider a triangle \(ABC\) with extended line \(CD\) forms an exterior angle to vertex \(C\).

**To prove**:

\(∠ACD = ∠A + ∠B\).

We will prove this using alternate angles property.

Let's draw a line \(CE\) from \(C\) which is parallel to \(AB\).

Take the line \(AC\) as transversal.

By alternate interior angle property, \(∠A = ∠ACD\) [alternate interior angles are equal in measure].

Now take the line \(BD\) as transversal for the parallel lines \(AB\) and \(CD\).

By corresponding angles property,\(∠B = ∠ECD\) [corresponding angles are equal in measure].

\(∠ACD = ∠ACE + ∠ECD = ∠A + ∠B\)

Thus, it is obvious that the exterior angle of a triangle is equal to the sum of its opposite interior angles.