Consider a general triangle ABC and the three inequalities given by the theorem.
Here, it will be shown that AC+AB>BC. The other two inequalities can be proved following the same procedure. Start by extending AB to the left of A. Then, consider a point D on this such that AD=AC.
In △ADC, the sides AD and AC are . This means that by the , the opposite them are . Therefore, ∠D≅∠DCA, which means that m∠D=∠DCA.
Notice that
∠DCB is made of
∠DCA and
∠ACB. Therefore, the measures of these three angles can be related thanks to the .
m∠DCB=m∠DCA+m∠ACB
From this ,
m∠DCB is greater than
m∠DCA.
m∠DCBm∠DCB>m∠DCA⇓>m∠D
Now consider
△DCB. From the previous inequality and the , the side opposite
∠DCB is longer than the side opposite
∠D.
m∠DCBDB>m∠D⇓>BC
The length of
DB is equal to the sum of the lengths of
DA and
AB. This is because of the . Substituting the corresponding sum into the left-hand side results in the following inequality.
DBDA+AB>BC⇓>BC
Recall that
D was plotted so that
AD=AC. Therefore, substitute
AC for
DA into the last inequality.
DA+ABAC+AB>BC⇓>BC✓
Notice that the last inequality is the desired one. Consequently, the proof is complete.