How do you prove that the exterior angle of a triangle is equal to the sum of the two opposite interior angles?

Answer 1

Assume a triangle ABC with a line AB (containing the side AB) with external angle D which is formed when line AB and line segment AC intersect. We are asked to prove that the external angle D is equal to the sum of the two interior angles B and C. Angles A and D are supplementary angles (they sum to 180 degrees) because they are linear angles (both together make a straight line, or a 180 degree angle). This means:

m<A + m<D = 180 degrees.

m<A = 180 deg - m<D

Then because A, B, and C are the three angles in a triange:

m<A + m<B + m<C = 180 deg

m<A = 180 deg - m<B - m<C

By substituting 180 deg - m<D in for m<A in the above equation we get:

180 deg - m<D = 180 deg - m<B - m<C

Subtract 180 deg from each side:

-m<D = -m<B - m<C

Multiply both sides by -1

m<D = m<B + m<C

Which proves that the measure of the external angle D is equal to the sum of the two opposite interior angles B and C for any given triangle.

wow. that's a lot. lol.