1
$\begingroup$

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:

enter image description 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:

enter image description here

How can I prove that the angles $A$ is the half of the difference of both angle measurements of Arc $BC$ and Arc $DE$?

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.