The perpendicular bisectors of sides AC and BC of ΔABC intersect side AB at points P and Q respectively, and intersect each other in the exterior (outside) of ΔABC. Find the measure of ∠ACB if m∠CPQ = 78° and m∠CQP = 62°.

The measure of ∠ACB will be 110°ExplanationAccording to the diagram below, [tex]DE[/tex] and [tex]DF[/tex] are the perpendicular bisectors of [tex]AC[/tex] and [tex]BC[/tex] respectively and they intersect side [tex]AB[/tex] at points [tex]P[/tex] and [tex]Q[/tex] respectively. So, [tex]AE=CE[/tex] and [tex]CF= BF[/tex]Now, according to the [tex]SAS[/tex] postulate, ΔAPE and ΔCPE are congruent each other. Also, ΔCFQ and ΔBFQ are congruent to each other. That means, ∠PCE = ∠PAE  and ∠FCQ = ∠FBQ As ∠CPQ = 78° , so  ∠PCE + ∠PAE = 78°  or,  ∠PCE = [tex]\frac{78}{2}= 39[/tex]°                                    and as ∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ = [tex]\frac{62}{2}=31[/tex]°Now, in triangle CPQ,  ∠PCQ = 180°-(78° + 62°) = 180° - 140° = 40°Thus, ∠ACB = ∠PCE + ∠PCQ + ∠FCQ = 39° + 40° + 31° = 110°