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

Answer: $\angle ACB=110^{\circ}$

Step-by-step explanation:

Here, 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 $\triangle ABC$

Let L is the mid point of side BC while M is the mid point of side AC.(mentioned on below figure).

Then, In triangles QBL and QLC,

QL=QL,  ( reflexive)

BL=LC, ( by the property of mid points)

And, $\angle QLB= \angle QLC$ ( right angles)

Therefore $\triangle QBL\cong \triangle QLC$, (SAS)

Thus, $\angle LQB=\angle LQC$ ( by CPCT)

But, $\angle LQB+\angle LQC+\angle CQP = 180^{\circ}$

Thus, $\angle LQB=59^{\circ}$

Now, In $\triangle QBL$,

$\angle QBL+\angle BLQ+\angle LQB=180^{\circ}$

⇒$\angle QBL=31^{\circ}$=$\angle ABC$

Similarly, $\triangle PMC\cong \triangle PMA$

Then $\angle MPC=\angle MPA$ ( by CPCT)

But, $\angle CPQ+\angle MPC+\angle MPA = 180^{\circ}$

Thus, $\angle MPA=51^{\circ}$

Now, In $\triangle MPA$,

$\angle MPA+\angle PAM+\angle AMP=180^{\circ}$

⇒$\angle MAP=39^{\circ}$=$\angle BAC$

Again, In $\triangle ABC$,

$\angle ABC+ \angle BAC+\angle ACB = 180^{\circ}$

$31^{\circ}+39^{\circ}+\angle ACB = 180^{\circ}$

⇒$\angle ACB=110^{\circ}$