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  $EF$  ?
 A: 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, $EF = BE - BF$.
A: 
$\dfrac{AB}{BC} =\dfrac{\sqrt3}{1} \implies \angle ACB=60^0$
$\triangle AFE$ is right angled. Use the fact that $AD$ is median of the triangle and get the side $FE$.
A: 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 $BE$.
see please picture

sorry  if picture is not good,i used paint
