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°.

Answer 1

Let M be the midpoint of side AC, then by SAS similarlity postulate, ΔAMP is congruent to ΔCMD.

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, and N be the midpoint of side BC, then by SAS similarlity postulate, ΔCNQ is congruent to ΔBNQ.

Thus, m<CQN = m<BQN

But, m<CQP + m<CQN + m<BQN = 180°
⇒ 2m<CQN = 180 - 62 = 118°
⇒ m<CQN = 59°

But, 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 + 40 + 31 = 110°