Equilateral triangle within a circumscribed circle
Andrew Mclaughlin 
In the attached picture there is an equilateral triangle within a circumscribed circle. MW is a radius of the circle.
I wish to prove that MT = TW, i.e., that the triangle's edge cuts the radius into equal parts.
I thought perhaps to draw lines AM and AW and to try and prove that I get two identical triangles, but failed to do so. Is it possible to prove this without trigonometry, using Euclidean geometry only ?
I need this because this is the basis for the second way to solve the Bertrand paradox in probability. While I'm OK in probability, I couldn't prove this crucial geometric aspect of the problem. Any help will be most appreciated here.
$\endgroup$4 Answers
$\begingroup$The centroid of every triangle lies on a median at $2/3$ of the way from the vertex to the opposite side. Hence $CM=2MT$ and you are done.
$\endgroup$ 2 $\begingroup$In right angled triangle $MBT$, $MB$ is a bisector.
Observe that $\angle BMW = 60°$, so $\dfrac{TM}{BM} = \cos60° = \dfrac12$, hence $MT = TW = \dfrac12 BM$
$\endgroup$ 2 $\begingroup$Join point $M$ to point $C$. By inscribed angle in same arc we have $\angle BCW= \angle BAW= 30°$ Similarly we have $\angle AWC= \angle ABC=60°$. In triangle $MAT$ we already know that $\angle MAT=30°$ and $\angle AMT=60°$. Hence by congruence in $ΔMAT$ and $ΔWAT$ we have $WT=MT$
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