Find the length of segment $EF$ in the figure below Right triangle $ABC$ has an altitude to the hypotenuse which is the diameter, $CD$ , of a circle. The circle intersects the triangle at $E$ and $F$ . Find the length of $EF$
$AC$ = 6 and $BC$ = 8
I found the length of $CD$ using trig but what is the approach to finding $EF$?
Here is an image:

 A: First, a more accurate diagram.

$ABC$ has area $\frac{1}{2}\cdot 6\cdot 8 = 24$.
$AB$ has length $\sqrt{6^2+8^2}=\sqrt{100}=10$.
$CD$ has length $24\cdot2\div10=4.8$.
$FG$ is the other diagonal of a rectangle with $CD$ and so has the same length, $4.8$.
A: ... please, draw a better figure next time...
Notice that $\angle ECF=90^\circ$ so we have $EF$ equals the diameter of the circle. Since $CD$ is another diameter, so $EF=CD$. Thus $EF=CD=\frac{6\times 8}{\sqrt{6^2+8^2}}=4.8$
A: Let make a code method by Mathematica for check each other
(scence = 
   RandomInstance[
    GeometricScene[{"A", "B", "C", "D", "E", "F", 
      "G"}, {Triangle[{"A", "B", "C"}], Triangle[{"E", "F", "C"}], 
      Line[{"A", "E", "C"}], Line[{"B", "F", "C"}], 
      PlanarAngle[{"A", "C", "B"}] == PlanarAngle[{"E", "C", "F"}] == 
       90 \[Degree], 
      GeometricAssertion[Line[{"B", "C"}], "Horizontal"], 
      GeometricAssertion[{CircleThrough[{"D", "E", "C", "F"}, "G"], 
        Line[{"A", "D", "B"}]}, {"Tangent", "D"}], 
      Line[{"D", "G", "C"}], EuclideanDistance["B", "C"] == 8, 
      EuclideanDistance["A", "C"] == 6, 
      Style[Line[{"E", "G", "F"}], Red]}], 
    RandomSeeding -> 3])["Graphics"]


EuclideanDistance["E", "F"] /. scence["Points"]


$4.8$

