Mickey Mouse birthday hat

Let $r$ and $s$ be the inradii of $\Delta ADF$ and $\Box AECF$ , and $2t$ be the side length of $\Box ABCD$ . Remember that the inradius of a polygon is one half of the ratio of its area and perimeter. Hence, $r=\frac{\frac{AD \cdot DF}{2}}{AD+DF+FA}=\frac{\frac{2t \cdot t}{2}}{2t+t+t\sqrt{5}}=(\frac{3-\sqrt{5}}{2})t \\ s=\frac{[ABCD]-[ADF]-[ABE]}{AF+FC+CE+EA}=\frac{4t^2-t^2-t^2}{t\sqrt{5}+t+t+t\sqrt{5}}=(\frac{\sqrt{5}-1}{2})t \\$ Now let $G$ be the foot of the perpendicular from $I_{2}$ to $DC$ ; $H$ and $I$ be the foot of the perpendiculars from $I_{1}$ to $AD$ and $I_{2}G$ , respectively. Now notice that $\frac{AH}{HI_{1}}=\frac{AD-DH}{HI_{1}}=\frac{2t-(\frac{3-\sqrt{5}}{2})t}{(\frac{3-\sqrt{5}}{2})t}=2+\sqrt{5}$ , and $\frac{I_{1}I}{II_{2}} = \frac{DC-r-s}{I_{2}G-r} = \frac{2t-(\frac{3-\sqrt{5}}{2})t-(\frac{\sqrt{5}-1}{2})t}{(\frac{\sqrt{5}-1}{2})t-(\frac{3-\sqrt{5}}{2})t} = 2+\sqrt{5} \\$ Thus, $\frac{AH}{HI_{1}} = \frac{I_{1}I}{II_{2}}$ , and so $\Delta AHI_{1} \sim \Delta I_{1}II_{2}$ . Since $\Delta AHI_{1}$ is right, therefore $90^\circ=\angle HAI_{1}+\angle AI_{1}H=\angle I_{2}I_{1}I+\angle AI_{1}H \\ \Rightarrow \angle AI_{1}I_{2}=\boxed{90^\circ}$