Exterior angle property of a triangle — lesson. Mathematics State Board, Class 7.

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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.

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3. Exterior angle property of a triangle