M=circumcenter of triangle ABC

den_thewhitelion wrote: M=circumcenter of triangle ABC Can you prove that?

Yes: MB=MC(triangle MBC has 2 20 angles) $\angle BMC=2\angle BAC$

where could you got the shortlist??

See here http://www.mathematica.gr/forum/viewtopic.php?f=58&t=49341

tenplusten wrote: den_thewhitelion wrote: M=circumcenter of triangle ABC Can you prove that? Because angle BMC=140 and we have that BAC=70 and also BM=MC

Note that $M$ is the circumcenter of triangle $ABC$. Since triangle $AMC$ is isosceles, $\angle MAB = 60^{\circ}$. Thus, since triangle $AMB$ is equilateral, $AB = BM = MC$, as desired.

Since $\angle ACM=10^\circ,$ we have that $\angle MBC=20^\circ.$ Hence, $\triangle MBC$ is isosceles with $MB=MC.$ Since $M$ lies in the interior of $\triangle ABC,$ it is the circumcenter of $\triangle ABC.$ Additionally, since $\triangle AMB$ has one angle of $60^\circ,$ we have that $\triangle AMB$ is equilateral. Hence, $AB=MB=MC=MA.$

sina/sin60=sin(70-a)/sin10=m/n we find a =60 and triangle is equilateral