Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$.

Question

Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$. If $DE = 15$ cm, $EF = 9$ cm, find $FG$.

What I Tried: Here is a picture :-

I have given variables for each of the sides, and there are a lot of things which I can do here.
I can use Pythagoras theorem on a lot of triangles, and more importantly, there are a whole lot of similar triangles.
For example, $\Delta DFC \sim \Delta GFB \sim \Delta GDA$ . From here :- $$\Rightarrow \frac{(z-a)}{z} = \frac{z}{(z + y)} = \frac{24}{(24 + x)}$$ And :- $$\Rightarrow \frac{z}{a} = \frac{(24 + x)}{x} = \frac{(z + y)}{y}$$ And :- $$\Rightarrow \frac{(z - a)}{a} = \frac{z}{y} = \frac{24}{x}$$ And using Pythagoras Theorem on multiple triangles gives even more equations. But I am confused here how to correctly use them and finally get the value of $x$, particularly because it is difficult for me to solve $4$ variables from here.

Can anyone help me? Thank You.

Answer 1

$CE$ is angle bisector. So use $$\dfrac{FE}{ED}=\dfrac{CF}{CD}=\dfrac{3}{5}$$

Thus $$\dfrac{FB}{CF}=\dfrac{2}{3}=\dfrac{FG}{FD}$$

Answer 2

Triangles $AED$ and $CEF$ are similar, and the ratio of their sides is $15:9=5:3$.
Hence this is the ratio of $AD:CF$.
Therefore $BF:CF=2:3$.

Triangles $ACF$, $GBF$ are also similar, with ratio $3:2$. Since $DF=24$, $FG=16$.

Answer 3

observe $\Delta DAE \sim \Delta FCE$

$ 15/9 = z/(z-a) $ so we got 'a' interms of 'z'

observe $\Delta GAD \sim \Delta GBF$

$ z/a = 24+x/x $ so done as we had gotten a in terms of z and we'll get $ x = 16 $