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Suppose that $ABC$ is a triangle with a median $AM$. Than we can look at these angles: $\angle ABC, \angle ACB, \angle BAM$ and $\angle CAM$. Is there any "nice relation" between them, so that I am able to express two of them in terms of the others? I have tried sine and Stewart's theorem for medians (and some other trigonometry theorems), but these techniques allow me to express only a jungle of trigonometric functions of these angles. Can anyone find anything better?

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  • $\begingroup$ In general, no. The law of cosines and/or Stewart's Theorem gets you all the information available -- and as you've seen, it's not pretty. $\endgroup$ – quasi Mar 13 '17 at 21:58
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$BA\cdot AM\cdot\sin\widehat{BAM} = AM\cdot AC\sin\widehat{CAM}$ (a median splits a triangle in two parts with the same area) hence $$\frac{\sin\widehat{BAM}}{\sin\widehat{CAM}}=\frac{b}{c}=\frac{\sin\widehat{ABC}}{\sin\widehat{ACB}}.$$

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  • $\begingroup$ Thanks very much. I hoped there is something not involving sines but I was wrong. Thank you for your effort. PS: It looks nice, indeed, even there are those sines. $\endgroup$ – Greenhorn3.14 Mar 15 '17 at 9:38

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