# a geometrical problem

$ABC$ is a right triangle with $∠ABC=90^0$ with $AB=30 \sqrt{3}$ and $BC=30$ . $D$ is a point on segment $B$C such that $AD$ is the median. $E$ is a point on segment $AC$ such that $BE$ is perpendicular to $AC$ . $AD$ and $BE$ intersect at $F$ . what is the value of $EF$ ?

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can you draw a figure? Geogebra is a nice tool fro drawing and you can then include it in your question. geogebra.org/cms – oks Apr 2 '13 at 9:26
i think it is triangle with angle $60,30,90$ – dato datuashvili Apr 2 '13 at 9:31

You can get $BE$ by taking the side $BC$ and $angle{EBC}$.

You can get $angle{EBC}$ knowing that $angle{EBF} = 60^0$

To get $BF$, you need to realise that you have $\angle{EBC}$, you can get $\angle{ADB}$ from the side $AB$ and $BD$ and then get $\angle{BFD}$, and last use the sine rule to get $BF$. From there, $EF = BE - BF$.

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thank you very much...i had got the answer..would you please tell me the answer so that i can check it... – sayan chaudhuri Apr 2 '13 at 9:49
Well, I think it would be better actually the other way round: you say what you got and we can check whether you got it right or wrong. – Jerry Apr 2 '13 at 15:35

$\dfrac{AB}{BC} =\dfrac{\sqrt3}{1} \implies \angle ACB=60^0$

$\triangle AFE$ is right angled. Use the fact that $AD$ is median of the triangle and get the side $FE$.

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more nice picture then mine :D – dato datuashvili Apr 2 '13 at 9:52

i think you should do following things 1.calculate value of $AC$,which is equal to $60$,so you see that at this right triangle ,you have following angles $30,90,60$ or $angle(BAC)=30$,$angle(ACB)=60$ and of course $angle(ABC)=90$ because $AD$ is median ,you can see that $BD=DC=15$

also you can find $BE$,you have $30$ degree angle and $BE$ is orthogonal to $AC$,or triangle $ABE$ is right triangle and use definition of oposite side from $30$ degree is half of hypotenus,now use following things find areas of triangles.this you need(maybe some other easy way will be shown) for finding heights,or for finding $BF$ and finaly substract $BF$ from $BE$.