Side BC of a Δ ABC is produced to both directions. Prove that the sum of the two exterior angles so formed is greater than 180° .

Given: ABC is a Δ and side BC is produced in both directions. Let CB be produced to P and BC be produced to Q.
To prove: ∠ ABP + ∠ ACQ > 180°
Proof:
When CB is produced to P,
Ext. ∠ ABP = ∠ A + ∠ C …………………(1)
(Ext. angle so formed will be equal to sum of the interior opposite angles)
When BC is produced to Q,
Ext. ∠ ACQ = ∠ A + ∠ B ……………….(2)
(Ext. angle so formed will be equal to sum of the interior opposite angles)
Adding (1) and (2), ∠ ABP+∠ ACQ = ∠ A+∠ C+∠ A+∠ B
                                              = ∠ A + (∠ A+∠ B+∠ C)
                                              = ∠ A + 180° [∠ A + ∠ B + ∠ C = 180° , since ∠ A, ∠ B and ∠ C are the three angles of a triangle .
∴ ∠ ABP + ∠ ACQ > 180°