# 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:

• Worth noting that this question is from ASMA (monthly short competition for middle/high school students). This round of the competition is now over. Commented Jan 26, 2022 at 0:11
• @Gust: CED is inscribed angle and CD is diameter $\Rightarrow$ CED = 90°. Also CFD = 90°, ECF = 90° $\Rightarrow$ CEDF is rectangle $\Rightarrow$ EF = CD. Commented Jan 26, 2022 at 11:56

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, $$4.8$$.

... 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 $$EF=CD=\frac{6\times 8}{\sqrt{6^2+8^2}}=4.8$$

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"]


$$4.8$$