In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$ , $BC = 100$ and $EC + ED = 150$, find $CF$.

Question

In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$, $BC = 100$ and $EC + ED = 150$, find $CF$.

What I Tried: Here is the diagram :-

I have assigned variables for the different lengths. From there we have $3$ equations :- $$x + y + a + z = 250$$ $$b + c = 100$$ $$y + z = 150$$ Which also implies :- $$x + a = 100$$ Now, $\Delta ABC \sim \Delta EFC$ . From here :- $$\frac{AC}{EC} = \frac{BC}{FC} = \frac{AB}{EF}$$ $$\rightarrow \frac{(x+y)}{y} = \frac{(b+c)}{c}$$ We also have $\Delta DBC \sim \Delta EBF$. From there we get :- $$\frac{DB}{EB} = \frac{CB}{FB} = \frac{CD}{EF}$$ $$\rightarrow \frac{(a+z)}{a} = \frac{(b+c)}{b}$$ $$\rightarrow \frac{250 - (x + y)}{a} = \frac{100}{100 - c}$$

After doing this I am completely lost and I am not finding any way to proceed. Can anyone help?

Answer 1

From $\triangle DEC \sim \triangle BEA$ $$\dfrac{DE}{CE}=\dfrac{BE}{AE}$$

$$\dfrac{150-y}{y}=\dfrac{100-x}{x}$$

$$\Rightarrow \dfrac{y}{x}=\dfrac{3}{2}$$

So $$CF=\dfrac{3}{3+2}BC=60$$

Answer 2

We also have,

$\frac{x}{y} = \frac{b}{c} = \frac{a}{z} = t$ (say)

So, $x + y + a + z = (y+z)(1+t) = 250$

We also have $b +c = c(1+t) = 100, y + z = 150$

Using $\, (y+z)(1+t) = 250, y + z = 150, c(1+t) = 100,$

$c = 60$.