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°.

The measure of ∠ACB will be 110°

Explanation

According to the diagram below, and are the perpendicular bisectors of and respectively and they intersect side at points and respectively.

So, and

Now, according to the 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 = °                                    and as ∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ = °

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

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

Step-by-step explanation:

Given :

The perpendicular bisectors of sides AC and BC of triangle ABC intersect side AB at points P and Q respectively and intersect each other in the exterior of triangle ABC.

AE = CE and CF = BF

Solution :

According to side angle side (SAS) postulate triangle APE and triangle CPE are congruent to each other. Therefore,

 ---- (1)

 ---- (2)

Now,

           (from (1))

Now,

           (from(2))

Now in triangle CPQ,

Now,

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