Question

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°.

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Answer 1

Let X be the midpoint of side AC, then by SAS similarlity postulate, ΔAXP is congruent to ΔCXD.
Thus, m∠APX = m∠CPX
But m∠APX + m∠CPX + m∠CPQ = 180° given that m∠CPQ = 78° 2m∠CPX = 180 - 78 = 102°m∠CPX = 102/2 =  51°
But m∠ACP = 90° - m∠CPX = 90 - 51 = 39°.
Similarly, and Y be the midpoint of side BC, then by SAS similarlity postulate, ΔCYQ is congruent to ΔBYQ.
Thus, m∠CQY = m∠BQY
But, m∠CQY + m∠CQP + m∠BQY = 180°⇒ 2m∠CQY = 180 - 62 = 118°⇒ m∠CQY = 118/2 = 59°
But, BCQ = 90 - m∠CQY = 90 - 59 = 31°
m∠PCQ = 180° - m∠CPQ - m∠CQP = 180 - 78 - 62 = 40°
Therefore, m∠ACB = m∠ACP + m∠PCQ + m∠BCQ = 39 + 40 + 31 = 110°Thus m∠ACB = 110°