Find the length of a segment in a triangle

In the drawing, if $$BF=3$$ and $$DE=4$$ and $$\angle BCD = 90º$$, find the lenght of $$CF$$

My try: I drew the segment $$BE=CE$$, because $$AE$$ is the angle bisector. After that i chased angles, and i found a lot of similar triangles, but after applying all the relations, i can't find the length of $$CF$$. I tried with sine and cosine law repeteadly, but it didn't work for me neither.

Any hints?

• I guess you have not mentioned that $\angle BCD$ is a right angle. – Lwins Jun 5 at 20:08
• @Lwins edited, thanks – Rodrigo Pizarro Jun 5 at 20:10
• And a quick observation: $BE \neq CE$ in general. – Lwins Jun 5 at 20:13

Since $$ACEB$$ is cyclic and $$CB||DE$$, we obtain: $$\measuredangle FCE=\measuredangle CED=x.$$ Thus, $$4=CE\cos{x}=CF\cos^2x=(BC-3)\cos^2x=(3\cot{x}\tan2x-3)\cos^2x.$$ Can you end it now?
I got $$CF=5.$$
• Did you mean $\measuredangle FCE=\measuredangle CED=x$? – Seyed Jun 5 at 20:38
$$\measuredangle FCE=\measuredangle CED=x.$$, From, $$\measuredangle ACB = 90 -2x$$ and $$\measuredangle ACE = 90-x$$ and since, $$\measuredangle ACB + \measuredangle FCE = \measuredangle ACE$$. So $$90- 2x + \measuredangle FCE = 90 -x$$ => $$\measuredangle FCE = x$$, $$\measuredangle CED =x$$, as it is alternate angle to $$\measuredangle FCE$$. $$\cos x = \frac{DE}{CE} = \frac{4}{CE} => CE = \frac{4}{\cos x}$$. Let $$y = BC - DE - BF => BC = 7 + y$$. So, from $$\triangle ACE$$, $$\sin x = \frac{CE}{AC}$$. From $$\triangle ABC$$, $$\sin 2x = \frac{BC}{AC}$$. Dividing, $$\frac{\sin 2x}{\sin x} = 2\cos x = \frac{BC}{CE} = \frac{7 + y}{CE}$$ Substituting for CE, $$2\cos x = \frac{7 + y}{CE} = \frac{7 + y}{\frac{4}{\cos x}} => 8 = 7 + y => y = 1 => CF = DE + y = 5$$