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

The measure of ∠ACB will be 110°

Explanation

According to the diagram below, $DE$ and $DF$ are the perpendicular bisectors of $AC$ and $BC$ respectively and they intersect side $AB$ at points $P$ and $Q$ respectively.

So, $AE=CE$ and $CF=BF$

Now, according to the $SAS$ 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 = $\frac{78}{2}=39$ °                                 and as ∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ = $\frac{62}{2}=31$°

Now, in triangle CPQ,  ∠PCQ = 180°-(78° + 62°) = 180° - 140° = 40°

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