Vertex $E$ of equilateral $\Delta ABE$ is in interior of $\square ABCD.$ $F$ is the midpoint of intersection of $AE$ and diagonal $BD$. If $AB = \sqrt{1 + \sqrt{3}}$ and $[\Delta ABF] = A$, find $(4A)^2$.
What I Tried: Here is a picture :-
From a bit of angle chasing, we have the angles of $\Delta ABF$ .
Also $AB = \sqrt{1 + \sqrt{3}}$, let $AF = y$. We have :-
$$\frac{y}{\sin B} = \frac{AB}{\sin C}$$
We have $\sin 75^\circ = \frac{1 + \sqrt{3}}{2\sqrt{2}}$ , so :-
$$\rightarrow y = \frac{\sqrt{1 + \sqrt{3}} * \sqrt{2}}{\frac{1 + \sqrt{3}}{2\sqrt{2}}*2}$$
Let $\sqrt{1 + \sqrt{3}} = k$. After continuous solving we will have :- $$y = \frac{2\sqrt{k}}{k}$$ From here I can get $$[\Delta ABF] = \frac{1}{2}(AB)(AF)\sin 60^\circ$$ $$\rightarrow \sqrt{k} * \frac{2\sqrt{k}}{k} * \frac{\sqrt{3}}{2}$$ $$\rightarrow [\Delta ABF] = \sqrt{3}$$
Hence $(4A)^2 = 48$ .
The question is, am I correct? Was it a coincidence that the $k$'s got cancelled out?
I am probably not correct, because right now I found out that $\sqrt{1 + \sqrt{3}} < \sqrt{3}$ , and this is impossible if my answer would have correct.
Can anyone let me know where I went wrong? Thank You.
Let's use $AB=k=\sqrt{1+\sqrt 3}$. Then $$\sin 75^\circ=\frac{1+\sqrt 3}{2\sqrt 2}=\frac {k^2}{2\sqrt 2}$$ Area of triangle $ABF$ is then: $$[\triangle ABF]=\frac12 AB\cdot AF\sin60^\circ=\frac12 k\frac {k\sin 45^\circ}{\sin 75^\circ}\sin 60^\circ\\=\frac 12 k^2\frac{\frac{\sqrt 2}2}{\frac{k^2}{2\sqrt 2}}\frac{\sqrt 3}2=\frac{\sqrt 3}2$$