From point $A$ lying outside a circle rays $AB$ and $AC$ come out and intersect the circle. Prove that the value of $\angle BAC$ is equal to half the difference of the angle measures of the arcs of the circle confined inside this angle.
I didn't understand the question well. Is this asking to prove $\frac{\mid\angle BOB_1 - \angle COC_1 \mid}{2} = \angle BAC$? How do I prove this?
$\qquad\qquad$
Let $O$ be the center of the circle. Also, let $B_1,B_2$ ($B_1$ is nearer to $A$) be the intersection points of the circle with $AB$ and $C_1,C_2$ ($C_1$ is nearer to $A$) be the intersection points of the circle with $AC$.
We want to prove that $$\frac{1}{2}(\alpha-\beta)=\gamma$$ where $\alpha=\angle{B_1OC_1},\beta=\angle{B_2OC_2}$ and $\gamma=\angle{BAC}$.
Let $\omega=\angle{B_1OB_2},\phi=\angle{C_1OC_2}$.
Using that the four triangles $OB_1B_2,OB_2C_2,OC_2C_1,OC_1B_1$ are isosceles and that the quadrilateral $B_1B_2C_2C_1$ is inscribed in the circle, we get $$\begin{align}\gamma&=180^\circ-\angle{AB_1C_1}-\angle{AC_1B_1}\\\\&=180^\circ-\angle{C_1C_2B_2}-\angle{B_1B_2C_2}\\\\&=180^\circ-(\angle{OC_2C_1}+\angle{OC_2B_2})-(\angle{OB_2B_1}+\angle{OB_2C_2})\\\\&=180^\circ-\left(\frac{180^\circ-\phi}{2}+\frac{180^\circ-\beta}{2}\right)-\left(\frac{180^\circ-\omega}{2}+\frac{180^\circ-\beta}{2}\right)\\\\&=-180^\circ+\frac{(\phi+\beta+\omega)+\beta}{2}\\\\&=-180^\circ+\frac{(360^\circ-\alpha)+\beta}{2}\qquad \text{(using $\alpha+\phi+\omega+\beta=360^\circ$)}\\\\&=\frac{1}{2}(\beta-\alpha)\end{align}$$
You can do similarly for the cases where there is a line tangent to the circle or $O$ is "outside" of $\angle{BAC}$.