Prove similarity of triangles

Question

I found this question in one of my textbooks.

In the given figure, ABCD and AEFG are 2 squares. Prove that:

$i) \frac{AF}{AG} = \frac{AC}{AD}$

$ii) ∆ACF \sim ∆ADG$

This is what I could do:

$i) AF^2 = AG^2 + GF^2\\ \rightarrow AF^2 = 2•AG^2 \rightarrow AF = AG•\sqrt{2} \rightarrow \frac{AF}{AG} = \sqrt{2}$

Similarly, $\frac{AC}{AD} = \sqrt{2}$

Therefore, $$\frac{AF}{AG} = \frac{AC}{AD}$$

In $(ii)$ part, $\frac{AF}{AG} = \frac{AC}{AD}$ can be one reason for similarity. What will be the other reason?

Answer 1

Reason 1 - $$\frac{AF}{AG} = \frac{AC}{AD}$$ Reason 2 - $\angle DAC = \angle GAF = 45°$ $\rightarrow \angle DAC - \angle GAC = \angle GAF - \angle GAC $ $$ \rightarrow \angle DAG = \angle CAF$$

$\therefore$ By $Side-Angle-Side$ congruency,

$$ \triangle ACF \sim \triangle ADG $$

Answer 2

Since a spiral similarity centred at $A$ maps $GF$ to $DC$, so there's also a spiral similarity (see for lemma #3 of this for a sketch of proof ) centred at $A$ mapping $GD$ to $FC$, hence QED.

Answer 3

Take $\Delta ADG$ and increase it at $\sqrt2$ such that you'll get $\Delta AD'G'$ with $D\in AD'$ and $G\in AG'.$

Now, rotate $\Delta AD'G'$ around $A$ on $45^{\circ}$.

You'll get $D'\rightarrow C$ and $G'\rightarrow F$ and we are done!

Maybe it will help to understand the Alex's reasoning.