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 $KM$.


enter image description here 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?

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    $\begingroup$ $AK=10-BK$ because $BKD$ is iscoseles triangle $\endgroup$ – MAGNUM Jan 20 at 17:08
  • $\begingroup$ Why the downvote? I added a diagram, wrote my thoughts on the problem. I think that this community really have too high expectations. Have a nice day! $\endgroup$ – Eager to learn math Jan 20 at 17:08
  • $\begingroup$ @MAGNUM, thank you! I solved it and got that $BK=7$. Can you give me a hint on finding $KM$? $\endgroup$ – Eager to learn math Jan 20 at 17:18
  • $\begingroup$ $KM=2KO, KO^2= BK^2-BO^2$ $\endgroup$ – MAGNUM Jan 20 at 17:23
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    $\begingroup$ They are alternate angles with respect to $AD // BC$. $\endgroup$ – player3236 Jan 20 at 17:44

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 $$x^2=(10-x)^2+64-2\cdot8\cdot(10-x)\cdot\dfrac12\Rightarrow x=7$$


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