Question

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Answer 1

Question 1

If we let $FE=x$, then $DE=x-5$.

Also, as $\overline{DF}$ bisects $\overline{BC}$, this means $BE=EC=6$.

Thus, by the intersecting chords theorem,

$6(6)=x(x-5)\\\\36=x^2 - 5x\\\\x^2 - 5x-36=0\\\\(x-9)(x+4)=0\\\\x=-4, 9$

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,

$2(32)=x^2\\\\64=x^2\\\\x=-8, 8$

However, as distance must be positive, we only consider the positive case, meaning CE=8