UPSKILL MATH PLUS

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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.