If one side of a triangle is longer than another side, then the measure of the angle opposite the longer side is greater than the measure of the angle opposite the shorter side.
Let's state exactly what is given and what we want to prove.
Given:& BC>AB, BD=BA
Prove:& m∠ BAC > m∠ C
Since the length of BD is equal to the length of BA, by the definition of congruent segments, BD is congruent to BA. Then, using the Base Angles Theorem, we can conclude that ∠ 1 is congruent to ∠ 2.
From the diagram, we can see that ∠ 2 is an exterior angle of △ ADC. Therefore, by the Triangle Exterior Angle Theorem, we know that m∠ 2 is the sum of m∠ 3 and m∠ C.
m∠ 2=m∠ 3+m∠ C
Let's now use the Transitive Property of Equality.
m∠ 2=m∠ BAC-m∠ 3 m∠ 2=m∠ 3+m∠ C
⇓
m∠ BAC-m∠ 3=m∠ 3+m∠ C
Finally, we will solve this equation for m∠ BAC.
As we can see, m∠ BAC is the sum of 2m∠ 3 and m∠ C. Note that all this measures are positive. Therefore, the obtained equation implies that m∠ BAC > m∠ C.
