Exterior Angle Theorem Math Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

hard
Prove that an exterior angle of a triangle is always greater than either of the two remote interior angles.

Solution

  1. 1
    Step 1: Let the exterior angle be ee and the remote interior angles be α\alpha and β\beta. By the Exterior Angle Theorem, e=α+βe = \alpha + \beta.
  2. 2
    Step 2: Since α\alpha and β\beta are interior angles of a triangle, both are positive (greater than 0°).
  3. 3
    Step 3: Therefore e=α+β>αe = \alpha + \beta > \alpha (since β>0\beta > 0) and e=α+β>βe = \alpha + \beta > \beta (since α>0\alpha > 0).
  4. 4
    Step 4: Thus the exterior angle is strictly greater than each remote interior angle.

Answer

Since e=α+βe = \alpha + \beta and both α,β>0°\alpha, \beta > 0°, we have e>αe > \alpha and e>βe > \beta.
This inequality follows directly from the Exterior Angle Theorem and the fact that all interior angles of a triangle are positive. Adding a positive quantity to α\alpha makes e>αe > \alpha, and similarly for β\beta. This result is useful in proofs about triangle geometry and the ordering of angles.

About Exterior Angle Theorem

An exterior angle of a triangle equals the sum of the two non-adjacent (remote) interior angles.

Learn more about Exterior Angle Theorem →

More Exterior Angle Theorem Examples