Theorem 2

Prove that: Prove that if two angles of a triangle are unequal in measure, the side opposite to the greater angle is longer than the side opposite to the smaller angle.
Given: In $\triangle ABC,$ $m\angle B$ > $m\angle C$
To prove: $m\overline{AC}$>$\overline{AB}$,
Construction: Make $\angle ABM \cong \angle C.$ Draw $\overline{BN}$, the bisector of $\angle MBC$, i.e. $m\angle 1=m\angle 2$.

Explanation

• To prove this theorem statement, we start by assuming that we have a triangle ABC with angle B greater than angle C.

• Then, we draw a line from point B to a point M on side AC, such that angle ABM is congruent to angle C. This creates two new angles, ABN and MBN, which we can use to prove our statement.

• We start by showing that angle ABN is congruent to angle ANB. This is because angle ANB is the sum of angles C and 2, while angle ABN is the sum of angles ABM and 1.

• We know that angles ABM and C are congruent because they are alternate interior angles formed by the transversal BM and the parallel lines AB and AC.

• We also know that angles 1 and 2 are congruent because they are both bisectors of angle MBC. Therefore, angles ABN and ANB are congruent by the angle addition postulate.

• Next, we use this congruence to show that side AC is longer than side AB. This is because if two angles of a triangle are congruent, then the sides opposite to those angles are congruent as well.

• In other words, AB is congruent to AN. Therefore, since angle ABN is congruent to angle ANB, we know that AB is congruent to AN. But AN is a segment of side AC, so we can conclude that AC is longer than AB.