First, consider a AD from A to the side BC so that BD is equal to AB. With this, an △ABD is created.
By the , it can be stated that ∠BAD and ∠BDA are .
Notice that, by the ,
m∠BAC can be written as the sum of
m∠BAD and
m∠DAC. m∠BAC=m∠BAD+m∠DAC
Since
∠BAD is congruent to
∠BDA, their measures are equal. Therefore,
m∠BDA can be substituted for
m∠BAD in this equation.
m∠BAC=m∠BAD+m∠DAC
m∠BAC=m∠BDA+m∠DAC
m∠BDA=m∠BAC−m∠DAC
Additionally, in
△ADC, note that
∠BDA is the to the non-adjacent angles
∠DAC and
∠C.
Therefore, by the , its measure is equal to the sum of the measures of
∠DAC and
∠C.
m∠BDA=m∠DAC+m∠C
Next,
m∠DAC+m∠C can be substituted for
m∠BDA into the previously obtained equation.
m∠BDA=m∠BAC−m∠DAC
m∠DAC+m∠C=m∠BAC−m∠DAC
m∠BAC=2m∠DAC+m∠C
Recall that the measure of every angle, including
∠DAC, is a value. Furthermore, observing the diagram, it can be seen that the measure of
∠DAC is
not 0∘. Under these conditions, it is implied that the measure of
∠BAC is greater than the measure of
∠C.
m∠BAC=2m∠DAC+m∠C⇓m∠BAC>m∠C
Therefore,
∠BAC, which is opposite the longer side
BC, is larger than
∠C, which is opposite the shorter side
AB. This concludes the proof.