Angles Inside the Circle Theorem

Let $E$ be the point of intersection of chords $\overline{AC}$ and $\overline{BD}.$ Start by drawing an auxiliary chord $\overline{BC}.$

Notice that $\angle 1$ is an exterior angle of $△BCE.$ Therefore, by the Triangle Exterior Angle Theorem, its measure equals the sum of the measures of the two non-adjacent interior angles. By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc. Finally, substituting these two equations into the equation given by the Triangle Exterior Angle Theorem, the first required equation is obtained.

To obtain the second equation, draw the auxiliary chord $\overline{AB}.$

As before, $\angle 2$ is an exterior angle of $△ABE.$ Therefore, its measure is equal to the sum of the measures of the two non-adjacent interior angles. Once more, the Inscribed Angle Theorem can be applied to rewrite the two angle measures on the right-hand side in terms of the corresponding intercepted arcs. Finally, substitute the last two equations into the one relating $m\angle 2$ and the measure of the inscribed angles.