# Proving the Secant Angles in the Circle

Ok, I know this is a very easy circle geometry problem, but I want to know that how to prove the theorem of angles in the circle. Like this image here: How can I prove that the angle $X$ is the half of the sum of both angles' measurement of the Arc $AC$ and Arc $BD$? This image here: How can I prove that the angles $A$ is the half of the difference of both angle measurements of Arc $BC$ and Arc $DE$?

For the second case.

Draw $\overline{BE}$ and let the measures of arcs $CB$ and $ED$ be $\alpha$ and $\beta$.

By the inscribed angle theorem, the measure of an angle inscribed in a circle is half the measure of its intercepted arc.

Therefore $\angle{E} = \frac{\alpha}{2}$

and $\angle{EBD} = \frac{\beta}{2}$

By the exterior angle theorem

$m \angle{A} + m \angle{E} = m \angle{EBD}$

Substituting

$m \angle{E} = \frac{\beta}{2} - \frac{\alpha}{2}$

For example, in the first case look at the triangle $\;\Delta BCK\;,\;\;K\;$ the point of intersection of both secants. Observe that

$$\angle x=180^\circ-\angle DCB-\angle ABC$$

But

$$\angle DCB=\frac12\angle BMC=\frac12\widehat{BC}\;,\;\;\angle ABC=\frac12\angle AMC=\widehat{AC}$$