This proof will consist of two parts. The first and second equations of the theorem will be proved one at a time.

$m\angle 1=\frac{1}{2}m\overgroup{AC}$

Consider a diameter $\overline{AF}.$ Since $\overleftrightarrow{AB}$ is tangent to the circle at $A,$ by the Tangent to Circle Theorem, $\overline{AF}$ and $\overleftrightarrow{AB}$ are perpendicular.

In addition, since $\overline{AF}$ is a diameter, $\overgroup{ACF}$ is a semicircle; its measure is therefore $180^{\circ }.$ By the Arc Addition Postulate, an equation that relates the measures of $\overgroup{AC}$ and $\overgroup{FC}$ can be written. Since $\angle CAF$ is an inscribed angle, by the Inscribed Angle Theorem, its measure is half the measure of the intercepted arc $\overgroup{FC}.$ Therefore, $2m\angle CAF$ can be substituted for $m\overgroup{FC}$ in the equation $m\overgroup{AC}+m\overgroup{FC}=180^{\circ }.$ Similarly, by the Angle Addition Postulate, it is evident that the sum of the measures of $\angle 1$ and $\angle CAF$ is $90^{\circ }.$ Next, multiply the equation above by $\text{-}2$ and add it to the $\text{previous}$ $\text{equation}.$ It is now possible to solve for $m\angle 1$ using this equation. The first equation of the theorem has been obtained.

$m\angle 2=\frac{1}{2}m\overgroup{ADC}$

The second equation of the theorem will now be obtained. To do so, the Arc Addition Postulate and the Angle Addition Postulate can be used to set the following pair of equations. Since $\overgroup{ADF}$ is a semicircle, its measure is $180^{\circ }.$ Additionally, since $\overline{FA}$ and $\overleftrightarrow{AB}$ are perpendicular, $\angle EAF$ is a $\text{right}$ $\text{angle}.$ Recall that it has been previously stated that $m\overgroup{FC}=2m\angle CAF.$ This information can be substituted in the above equations. Next, multiply Equation $(\text{II})$ by $\text{-}2$ and add it to Equation $(\text{I}).$ Finally, by solving the resulting equation for $m\angle 2,$ the second equation of the theorem is obtained.