Question

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

Answer 1

∠ABC will be equal to 110°

In this DE and DF are the bisectors which are perpendicular to AC and BC, and they also intersect side AB at point P and Q.

So we can get this from the above statement:

AE = CE and CF = BF

ΔAPE and ΔCPE are congruent each other.

And

ΔCFQ and ΔBFQ are congruent to each other.

∠PCE = ∠PAE  and ∠FCQ = ∠FBQ

∠CPQ = 78° , so  ∠PCE + ∠PAE = 78°  or,  ∠PCE = 78/2 = 39°

∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ = 62/2 = 31°

∠CQP, ∠PCQ = 180°-(78° + 62°) = 180° - 140° = 40°

∠ACB = ∠PCE + ∠PCQ + ∠FCQ = 39° + 40° + 31° = 110°