Find the area of triangle ABC given areas of smaller triangles 
Consider a point P in the interior of a triangle ABC. Draw three lines passing through the point and parallel to the sides of the triangle.

They divide the triangle into three parallelograms and three triangles. Let S1, S2, and S3 be the areas of these smaller triangles. Find the area of the triangle ABC.

This is what I have completed so far but I am really stuck.

 A: Let $S$ be area of $\triangle ABC$.
Let $(u,v,w)$ be barycentric coordinate of point $P$ with respect to $\triangle ABC$.
By definition, $u,v,w$ are three numbers such that
$$u+v+w = 1\quad\text{ and }\quad\vec{P} = u\vec{A} + v\vec{B} + w\vec{C}$$
Since $P$ lies in the interior of $\triangle ABC$, $u, v, w > 0$.
Furthermore, the barycentric coordinates are ratio of heights of triangles. As an example,
$$v = 
\text{ height of } \triangle PAC \text{ at } P : 
\text{ height of } \triangle ABC \text{ at } B
$$
Since $\triangle PIE$ is similar to $\triangle ABC$ and has same height as $\triangle PAC$ at $P$, we find
$$S_3 = \verb/Area/(\triangle PIE) = \verb/Area/(\triangle ABC) v^2 = S v^2$$
By a similar argument, we have $S_2 = S u^2$ and $S_1 = Sw^2$.
Since $u, v, w > 0$, we obtain
$$\sqrt{S} = \sqrt{S}(u+v+w) = \sqrt{Su^2} + \sqrt{Sv^2} + \sqrt{Sw^2} = \sqrt{S_2} + \sqrt{S_3} + \sqrt{S_1}$$
A: 
Similar triangles are to one another in the squared ratio of (their) corresponding sides[1]
Therefore, $$\small [\triangle FIP]:[\triangle PHE]:[\triangle GPG]=S_1:S_2:S_3$$ $$\small\implies FP:PE:GJ=\sqrt{S_1}:\sqrt{S_2}:\sqrt{S_3}$$
From that $\small AC:GJ=(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3}):\sqrt{S_3}$
As $\small \triangle ABC\sim\triangle GPJ\implies$ $$\small \begin{align} \small \frac{[\triangle ABC]}{[\triangle GPJ]}&=\frac{(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2}{(\sqrt{S_3})^2}\\ \small \frac{[\triangle ABC]}{S_3}&=\frac{(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2}{{S_3}}\\ \boldsymbol {[\triangle ABC]}&\boldsymbol{=(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2}\end{align}$$
