SOLUTION: 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 &#91

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Question 1129840: 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
Answer by MathLover1(20850) About Me  (Show Source):
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Let M be the midpoint of side AC, then by SAS similarity postulate, Δ AMP+is congruent to Δ CMD.
Thus, m < APM = m < CPM
But m < APM++ m < CPM + m < CPQ = 180°
⇒ 2m < CPM+= 180+-+78+=+102°
⇒ m < CPM+=+51°
But m < ACP+= 90° - m < CPM = 90+-+51+=+39°.
Similarly, let N be the midpoint of side BC, then by SAS similarity postulate, Δ CNQ is congruent to Δ BNQ.
Thus, m < CQN =+m < BQN
But, m < CQP + m < CQN + m < BQN+=+180°
⇒ 2m < CQN+= 180+-+62+=+118°
⇒ m < CQN+=+59°

But, m <BCQ = 90 - m < CQN+= 90+-+59+=+31°
m < PCQ = 180° -+m < CPQ - m < CQP = 180+-+78+-+62+=+40°
Therefore, m < ACB = m < ACP++ m < PCQ++ m < BCQ+= 39+%2B+40+%2B+31+=+110°


so, your answer is:

m < ACB =110°