Let $A B C$ be an isosceles triangle with $A B=A C$ and $\widehat{A}=30^{\circ}$. Let $D$ be the midpoint of the base $B C$. On $A D$ and $A B$, take two points $P$ and $Q$, respectively, such that $\mathrm{PB}=\mathrm{PQ}$. Determine the measure of the angle $\angle PQC$.
Let us observe that

$$
A \hat{B C}=A \hat{C} B=\frac{180^{\circ}-30^{\circ}}{2}=75^{\circ}
$$

Since all points on the altitude AP are equidistant from B and C, in particular, triangle BPC is isosceles with $B P=P C$. By the problem's hypothesis, triangle BPQ is also isosceles. We denote by $\alpha$ the measure of angle $\mathrm{PBC}$, so $\mathrm{BCP}=\alpha$ and

$$
\hat{A Q P}=180^{\circ}-\mathrm{BQP}=180^{\circ}-\mathrm{QBP}=180^{\circ}-\left(75^{\circ}-\alpha\right)=105^{\circ}-\alpha
$$

and

$$
P \hat{C A}=75^{\circ}-P \hat{C} B=75^{\circ}-\alpha
$$

Thus $A \hat{Q} P+P \hat{C A}=180^{\circ}$, therefore quadrilateral $A Q P C$ is cyclic, in particular $\mathrm{PQC}=\mathrm{PA} \mathrm{C}=15^{\circ}$.