94. The Exterior Angle Theorem · The outside angle equals the two inside opposites

Try to break it

Try moving A so the triangle gets very tall or very flat. Does the relationship still hold? What happens if you drag A below the line BC?

How you build it

Make a triangle with an extended side.

• Place point A, the apex of the triangle.
• Place point B, the left end of the base.
• Place point C, the third vertex, to the right of B.
• Draw a ray starting at B, through C and beyond.
• Draw the side from A to B.
• Draw the side from C to A to complete triangle ABC.
• Place point D on the ray, beyond C.

The proof, step by step

Prove that exterior angle ∠ACD equals the sum of interior angles ∠A and ∠B.

• We know the angles in triangle ABC add up to 180°: ∠A + ∠B + ∠BCA = 180°.
• Angles on a straight line add up to 180°: ∠BCA + ∠ACD = 180°.
• Therefore, ∠A + ∠B + ∠BCA = ∠BCA + ∠ACD.
• Subtract ∠BCA from both sides to get ∠ACD = ∠A + ∠B.

Worked example

In triangle PQR, side QR is extended to S. If ∠P = 55° and ∠Q = 65°, what is the measure of the exterior angle ∠PRS?

By the Exterior Angle Theorem, the exterior angle ∠PRS equals the sum of the two opposite interior angles: ∠P + ∠Q = 55° + 65° = 120°.

• 110°
• 120° — correct
• 130°
• 100°