Awesome geometry - 10

Let

$\overline{FE} = a$

$\overline{DE} = b$

$\overline{BE} = c$

$\overline{BD} = d = 12$

$\overline{DC} = e = 18$

$\overline{BC} = f$

We have a theorem related to tangents -

$f^2 = e \times (a + b + e) \dots\ (1)$

Also Pythagoras theorem gives us-

$f^2 = c^2 + (b + e)^2 \dots\ (2)$

Equating $(1) , (2)$ we get ,

$e \times (a + b + e) = c^2 + (b + e)^2$

$e(a + b) + e^2 = c^2 + b^2 + 2.b.e + e^2$

$e(a + b) = c^2 + b^2 + 2.b.e$

Dividing both sides by $e$ ,

$a + b = \frac{c^2 + b^2}{e} + 2b$

Also , pythagoras theorem gives us , $c^2 + b^2 = d^2$

$(a + b) - 2b = \frac{d^2}{e}$

$(a + b) - 2b = \frac{12^2}{18}$

$(a + b) - 2b = \frac{144}{18}$

$(a + b) - 2b = 8$

$\overline{DF} - 2 \times \overline{DE} = \boxed{8}$

Imgur

$c^2 = 144-a^2$

From intersecting secant theorem: $CF \times CD = CB^2$

$(18+a+b)(18)=t^2=c^2+(a+18)^2$

$324+18a+18b=144-a^2+a^2+36a+324$

divide by 18 throughout and cancelling terms $a+b = 8 + 2a$

$b - a = 8$