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 $DE$?
$\endgroup$2 Answers
$\begingroup$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
$m \angle{E} = \frac{\beta}{2} - \frac{\alpha}{2}$
$\endgroup$ $\begingroup$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
$$\angle DCB=\frac12\angle BMC=\frac12\widehat{BC}\;,\;\;\angle ABC=\frac12\angle AMC=\widehat{AC}$$
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