Question 1
If we let [tex]FE=x[/tex], then [tex]DE=x-5[/tex].
Also, as [tex]\overline{DF}[/tex] bisects [tex]\overline{BC}[/tex], this means [tex]BE=EC=6[/tex].
Thus, by the intersecting chords theorem,
[tex]6(6)=x(x-5)\\\\36=x^2 - 5x\\\\x^2 - 5x-36=0\\\\(x-9)(x+4)=0\\\\x=-4, 9[/tex]
However, as distance must be positive, we only consider the positive case, meaning FE=9
Question 2
If we let CE=x, then because AB bisects CD, CE=ED=x.
We also know that since FB=17, the radius of the circle is 17. So, this means that the diameter is 34, and as AE=2, thus means EB=32.
By the intersecting chords theorem,
[tex]2(32)=x^2\\\\64=x^2\\\\x=-8, 8[/tex]
However, as distance must be positive, we only consider the positive case, meaning CE=8
