The outer angle at any vertex of a triangle is equal to the sum of the two inner angles at the other vertices. This is proven using the properties that alternate angles are equal and corresponding angles are equal when two parallel lines are cut by a transversal. Specifically, if angle CBD is the outer angle at vertex B, it is shown to be equal to the sum of the inner angles ACB and CAB by using alternate and corresponding angles formed by drawing a line parallel to one side of the triangle.

1. Outer Angle
Theorem

2. Alternate angles

3. Corresponding Angles

4. The outer angle at a vertex is
equal to the sum of the inner
angles at the other two vertices

12. Prove that in any triangle, the outer angle at a
vertex is equal to the sum of the inner angles at
the other two vertices?
Consider the triangle ABC.
< CBD is the outer angle.
We have to prove that <CBD=<ACB+<CAB

13. Draw BE parallel to AC

14. AC parallel to BE
BC is transversal
<ACB=<CBE (Alternate angles)
AB is transversal
<CAB=<DBE (corresponding angles)

15. <ACB=<CBE
<CAB=<DBE
From 1,
<ACB+<CAB=<CBE+<DBE
<ACB+<CAB=<CBD
1
Therefore, outer angle at a vertex is equal to
the sum of the inner angles at the other two
vertices