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?
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 $CF=5.$
$\measuredangle FCE=\measuredangle CED=x.$, 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, $$2\cos x = \frac{7 + y}{CE} = \frac{7 + y}{\frac{4}{\cos x}} => 8 = 7 + y => y = 1 => CF = DE + y = 5$$
