The parallelogram $ABCD$ is determined by $AB=8,AD=10$ and $\measuredangle BAD=60^\circ$. The perpendicular bisector $s_{BD}$ of $BD$ intersects $AD$ and $BC$ at $K$ and $M$, respectively. Find $BK$ and $KM$.
$BK=7,KM=4\sqrt7$
First, I am trying to find $BK$. The cosine rule on triangle $ABK$ gives $$BK^2={AK}^2+AB^2-2.AK.AB.\cos60^\circ.$$ We don't know only the length of $AK$. How can I find it?
Expanding on Magnum's comment. As you pointed out $$BK^2=AK^2+AB^2-2\cdot AK\cdot AB\cdot\cos\measuredangle BAK$$ Note that $BK=DK$ so let $BK=DK=x$. Then $AK=10-x$. Plugging in gives $$x^2=(10-x)^2+64-2\cdot8\cdot(10-x)\cdot\dfrac12\Rightarrow x=7$$