Let $BM$ be the median of triangle $ABC$. On the extension of $MB$ beyond $B$, the point $K$ is chosen so that $BK =\frac12 AC$. Prove that if $\angle AMB=60^o$, then $AK=BC$.
(Mykhailo Standenko)
Let $AC =2k$
$AB=p$
$BC=q$
Stewart
$BM = \sqrt{\frac{p^2+q^2}{2}-k^2}$
Cosine Rule On BC we get this relation
$\frac{(q^2-p^2)}{2} = \sqrt{\frac{p^2+q^2}{2}-k^2}*k$
Cosine rule on AK
$AK^2 = k^2+(\sqrt{\frac{p^2+q^2}{2}-k^2}+k)^2-k*(\sqrt{\frac{p^2+q^2}{2}-k^2}+k)$
$AK^2 = k^2 +(\frac{p^2+q^2}{2}-k^2+k^2+2k\sqrt{\frac{p^2+q^2}{2}-k^2})-k^2-\frac{(q^2-p^2)}{2}$
$AK^2 = k^2 +(\frac{p^2+q^2}{2}+(q^2-p^2))-k^2-\frac{(q^2-p^2)}{2}$
$AK^2 = \frac{(p^2+q^2)}{2}+\frac{(q^2-p^2)}{2}$
$AK^2=q^2=BC^2$
$AK=BC$
Take $T$ such that $BMAT$ is a parallelogram ($TM$ diagonal). $KB=BT$ and $KBT=60$ hence $KBT$ is equilateral and $KT=MA$ which along with $BM\|TA$ means $KMAT$ is a isosceles trapezoid. Hence $AK=TM=BC$.