# Parallelogram and cosine rule

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?

• $AK=10-BK$ because $BKD$ is iscoseles triangle – MAGNUM Jan 20 at 17:08
• 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! – Eager to learn math Jan 20 at 17:08
• @MAGNUM, thank you! I solved it and got that $BK=7$. Can you give me a hint on finding $KM$? – Eager to learn math Jan 20 at 17:18
• $KM=2KO, KO^2= BK^2-BO^2$ – MAGNUM Jan 20 at 17:23
• They are alternate angles with respect to $AD // BC$. – 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 $$x^2=(10-x)^2+64-2\cdot8\cdot(10-x)\cdot\dfrac12\Rightarrow x=7$$