Nasty Triangles 3

$\angle ABC = \angle ACB, \angle ADE = 180 - \angle ABC \Rightarrow \angle CDE=\angle ABC \Rightarrow \triangle ABC$ and $\triangle DCE$ are similar $\Rightarrow DE = EC, BC =BE + DE$ and $\frac{3}{BC}=\frac{DE}{2} \Rightarrow DE \times BC=6$ $ABED$ is cyclic $\Rightarrow$ $\frac{AE}{DB} = \frac{AB \times AD + BE \times DE}{AB \times BE + AD \times DE} = \frac{\sqrt{5}}{2}$ From these we get $DE = \sqrt{5}, BE = \frac{1}{\sqrt{5}}$ Using Menelaus Theorem: $\frac{AF}{AF-3}\frac{1}{5}\frac{2}{1}=1$ we get $AF=5$ Using Menelaus Theorem again: $\frac{3}{2}\frac{DE}{EF}\frac{2}{3}=1$ we get $DE=EF=\sqrt{5}=EC$ . From this we get $\angle ACF = 90$ . $CF=\sqrt{5^2-3^2}=\boxed{4}$

Suppose $BE=a$ , $EC=b$ , $AE=c$ and $BD=\frac{2}{\sqrt5}c$ , because $AB=AC=3$ , $\angle ABC=\angle ACB=\angle ECD$ , so $DE=EC=b$ . According to Stewart's theorem , $AE^2=\frac{AB^2\times EC+AC^2\times BE}{BE+EC}-BE\times EC$ $\begin{aligned} c^2&=\frac{3^2 b+3^2 a}{a+b}-ab \\&=9-ab \end{aligned}$ $ab=9-c^2$ According to secant-secant theorem , $CD\times CA=CE\times CB$ $b(b+a)=2\times 3$ $ab+b^2=6$ According to Ptolemy's theorem , $AB\times DE+AD\times BE=AE\times BD$ $3b+a=\frac{2}{\sqrt5}c^2$ Now we have $\begin{cases} ab=9-c^2 \\ ab+b^2=6 \\ 3b+a=\frac{2}{\sqrt5}c^2 \end{cases}$ Solve the system of equations above and you would get $a=\frac{1}{\sqrt5}$ , $b=\sqrt5$ , $c=\sqrt8$ .

Now according to Menelaus' theorem , $\frac{CD}{DA}\times \frac{AF}{FB}\times \frac{BE}{EC}=1$ $\frac{2}{1}\times \frac{3+BF}{BF}\times \frac{\frac{1}{\sqrt5}}{\sqrt5}=1$ So, $BF=2$ , according to Menelaus' theorem again, $\frac{AB}{BF}\times \frac{FE}{ED}\times \frac{DC}{CA}=1$ $\frac{3}{2}\times \frac{FE}{\sqrt5}\times \frac{2}{3}=1$ Hence, $FE=\sqrt5$ , notice $FE=ED=EC=\sqrt5$ and $FED$ is a straight line, we can conclude $\angle FCD=90^\circ$ .

Thus, $\begin{aligned} CF&=\sqrt{AF^2-AC^2} \\&=\sqrt{5^2-3^2}\\&=\sqrt{25-9}\\&=\sqrt{16}\\&=4 \end{aligned}$