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In the attached picture there is an equilateral triangle within a circumscribed circle. MW is a radius of the circle.

enter image description here

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.

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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.

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  • $\begingroup$ both radius and then I'm done...I see. I didn't figure out that it's the median, I thought it was the bisection (all bisections meet at the center of the triangle). $\endgroup$ – Jankel Jan 19 '18 at 16:36
  • $\begingroup$ In an equilateral triangle, medians are the same as perpendicular bisectors. $\endgroup$ – Aretino Jan 19 '18 at 17:42
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In right angled triangle $MBT$, $MB$ is a bisector. enter image description here

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Observe that $\angle BMW = 60°$, so $\dfrac{TM}{BM} = \cos60° = \dfrac12$, hence $MT = TW = \dfrac12 BM$

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  • $\begingroup$ Why is the angle being 60 degrees ? $\endgroup$ – Jankel Jan 19 '18 at 16:31
  • $\begingroup$ By symmetry, $\angle AMB = 360°/3 = 120°$, and $\angle BMW = \angle AMW = 120°/2 = 60°$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 19 '18 at 16:38
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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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