<s>[INST] Geometry : what is the $\phi$ angle, if area of yellow rectangle is equal with area of red triangle? I have a right triangle and in it area of yellow rectangle is equal with area of red triangle. How could prove that $\phi=45^{\circ}$?
$$\text{Area of  Yellow  Rectangle}=\text{Area  of Red  Triangle}$$

This problem has no additional information.
I'm sorry for the inconvenience.
Update:
If we add some constraint to this problem like bellow, is it have a unique solution?
$$|FG|=x+y$$

 [/INST] Thanks for all. 
first equality: area of red and yellow are equal.
$$ 2xy=vw \label{eq:1} \tag{1}$$
Now by using Thales' theorem in this triangle:
$$ \frac{v+y}{c}=\frac{b-x}{b} \Rightarrow v=c(\frac{b-x}{b})-y       \label{eq:2} \tag{2}$$
$$ \frac{x+w}{b}=\frac{c-y}{c} \Rightarrow w=b(\frac{c-y}{c})-x       \label{eq:3}   \tag{3}$$
by importing $\ref{eq:2}$ and $\ref{eq:3}$ in $\ref{eq:1}$ and simplifying equations, I gained this equation:
$$ \frac{(x-b)^2}{b^2}+\frac{(y-c)^2}{c^2}=1 \label{eq:4}   \tag{4}$$
It's means location of P, which area of red and yellow are equal, is an elliptic with canonical point $B$ and $C$. So we have $\infty$ points for $P$ in the $ABC$.
Now, How to find $\phi$?

By loking at this picture we have:
$$ AG=\sqrt{(x+w)^2+y^2} \label{eq:5}   \tag{5}$$
$$ AF=\sqrt{(y+v)^2+x^2} \label{eq:6}   \tag{6}$$
$$ FG=\sqrt{v^2+w^2} \label{eq:7}   \tag{7}$$
Law of cosines
$$ FG^2=AG^2+AF^2-2AF.FG.\cos(\phi) \label{eq:8}   \tag{8}$$
Now by using $\ref{eq:2}$ and $\ref{eq:3}$ in $\ref{eq:5}$, $\ref{eq:6}$ and $\ref{eq:7}$ and finally in $\ref{eq:8}$, and simplifying that, I reach to this equation:
$$ \sqrt{\left[\frac{b^2}{c^2}(c-y)^2+y^2 \right]\left[\frac{c^2}{b^2}(b-x)^2+x^2 \right]}\cos(\phi)= \left[bx+cy-xy(\frac{b^2+c^2}{bc})\right]\label{eq:9}   \tag{9}$$
So 
$$ \cos(\phi)=\frac{\left[bx+cy-xy(\frac{b^2+c^2}{bc})\right]}{\sqrt{\left[\frac{b^2}{c^2}(c-y)^2+y^2 \right]\left[\frac{c^2}{b^2}(b-x)^2+x^2 \right]}}\label{eq:10}   \tag{10}$$
$$for \: b > x > 0 \: and \: c > y > 0 \: $$
For finding a $\phi$ as a function on $x$ and $y$ we need to considering $\ref{eq:4}$ and $\ref{eq:9}$. If I find exact solution I post it here.
I have writed a python code for determine all possible solution of this equation, but this not have a unique solution. Since all $x$ and $y$ that for them we have $\cos(\phi)=\frac{\sqrt{2}}{2}$ are solutions.
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