I need a short way to solve following problem:

Suppose that length of chord $AB=5\operatorname{cm}$,$AC=7\operatorname{cm}$ and $BC=8\operatorname{cm}$, we know that $D$ is midpoint of arc $BC$ and chord $AD$ divides $BC$ into two equal parts (let intersection point be $K$) so $BK=KC=4$, we are going to find $AK$ and $KD$.

Problem illustration

First of all I know characters of chords intersection , which means that $AK\cdot KD=BK \cdot KC$, sure we can find by cosine theorem $AK$(we know all length,we can find any angle and then repeat usage of cosine)but because $D$ is midpoint of arc $BC$ and also $AD$ is median, I doubt that AD is the bisector, diameter or something like that.

Thanks a lot.

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    $\begingroup$ It would be nice in problems like this to draw a diagram. Even a paint program will make one good enough for the purpose. $\endgroup$ – Ross Millikan Apr 4 '12 at 14:07
  • $\begingroup$ sorry for my bad pointing,but is is like this $\endgroup$ – dato datuashvili Apr 4 '12 at 14:26
  • $\begingroup$ no 25+49!=64,no diameter $\endgroup$ – dato datuashvili Apr 4 '12 at 14:28
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    $\begingroup$ @SalechAlhasov: $5^2+7^2=25+49=74 \ne 8^2$ $\endgroup$ – Ross Millikan Apr 4 '12 at 14:28
  • $\begingroup$ but actually cosine theorem does not work here,i did not get correct answer ,why i can not see $\endgroup$ – dato datuashvili Apr 4 '12 at 14:54

I get consistency with the cosine theorem. $7^2=8^2+5^2-2\cdot 8 \cdot 5 \cos B$ leads to $\cos B= \frac 12$ and $AK=\sqrt {21}$. Similarly $\cos C=\frac {88}{112}$ and again $AK=\sqrt {21}$



For a double check that $AD \neq$ radius, the circumradius of $ABC$ is $r =\dfrac{7}{\sqrt{3}}.$ You probably made a mistake in calculations using cosine.

  1. The sides of $ABC$ are $5,7,8.$ So the median (wikipedia) $AK$ is $\sqrt{21}.$

  2. Now that you have $KD,$ use the the equation in your question.


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