72. Let $P$ and $Q$ be two distinct points on the circumcircle of triangle $ABC$ such that $|PA|^2 = |PB| \cdot |PC|$, $|QA|^2 = |QB| \cdot |QC|$ (one of the points is on the arc $\breve{AB}$, the other is on the arc $\widehat{AC}$). Find the difference $\widehat{PAB} - \widehat{QAC}$, if the difference of the angles $\hat{B}$ and $\hat{C}$ of triangle $ABC$ is $\alpha$.
72. If $P$ lies on the arc $\overrightarrow{A B}$, and $Q$ lies on the arc $\overrightarrow{A C}$, then for the angles $\widehat{P A B}=\dot{\varphi}, \widehat{Q A C}=\psi$ we obtain two relations:

$$
\begin{aligned}
&\left\{\begin{aligned}
\sin ^{2}(\hat{C}-\varphi) & =\sin \varphi \sin (\hat{B}+\hat{C}-\varphi) \\
\sin ^{2}(\hat{B}-\psi) & =\sin \psi \sin (\hat{B}+\hat{C}-\psi)
\end{aligned}\right. \\
& \Rightarrow\left\{\begin{array}{l}
1-\cos 2(\hat{C}-\varphi)=\cos (\hat{B}+\hat{C}-2 \varphi)-\cos (\hat{B}+\hat{C}) \\
1-\cos 2(\hat{B}-\psi)=\cos (\hat{B}+\hat{C}-2 \psi)-\cos (\hat{B}+\hat{C})
\end{array}\right.
\end{aligned}
$$

Let's write the difference of these equations:

$\sin (\hat{B}+\hat{C}-\varphi-\psi) \sin [(\hat{B}-\hat{C})+(\varphi-\psi)]=$

$=\sin (\hat{B}+\hat{C}-\varphi-\psi) \sin (\varphi-\psi)$,

from which (since $0<\hat{B}+\hat{C}-\varphi-\psi<\pi$) we get that $\hat{B}-\hat{C}+$ $+(\varphi-\psi)=\pi-(\varphi-\psi)$.

Answer: $\frac{\pi-\alpha}{2}$.