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.
Best Answer
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 $DF=24$, $FG=16$.