area of il-defined triangle $AA'$, $BB'$, and $CC'$ are straight lines drawn from the angular points of a triangle through any point $O$ within the triangle and cutting the opposite sides at $A'$, $B'$, and $C'$. $AP$, $BQ$, and $CR$ are cut off from $AA'$, $BB'$, and $CC'$ and are equal to $OA'$, $OB'$, and $OC'$.
Prove that the area of triangle $\Delta PQR$ equals area of triangle $\Delta A'B'C'$.

Context: early 1900 school book based on Euclid. At the end of Book II  [pythagorus, length of side opposite angle, area of triangle = SQRT(s(s-a)(s-b)(s - c))] the question:
 A: Let $(x,y,z)$ be the barycentric coordinates of the point $O$ w.r.t. the given triangle $\Delta ABC$. They can be defined as
$$
\begin{aligned}
x &= OA':AA'\ ,\\
y &= OB':BB'\ ,\\
z &= OC':CC'\ ,\\
1 &= x+y+z\ ,
\end{aligned}
$$
as suggested in the following picture:

We can then write $O=xA+yB+zC$.
For a quick guide i often use Barycentric coordinates for the impatient, Max Schindler, Evan Chen.

First solution, we blindly use barycentric coordinates:
Let us compute $P$. We have $A(1,0,0)$ and 
$A'(0:y:z)=(0,\frac y{y+z},\frac z{y+z})$, the mid point of $AA'$ is
$M=\frac 12(A+A')
%=\left(\frac 12,\frac y{2(y+z)},\frac z{2(y+z)}\right)
$, and finally $P$ is the reflection of $O$ in $M$, so 
$$
P=2M-O=A+A'-O
=
\left(1+0-x,\ 0+\frac y{y+z}-y,\ 0+\frac z{y+z}-z\right)
=
\left(1-x,\ \frac {xy}{y+z},\ \frac {xz}{y+z}\right)
\ .
$$
In this situation, the points $A'$, $B'$, $C'$; $P$, $Q$, $R$ have the coordinates
$$
\begin{aligned}
A' &= (0:y:z)=\frac 1{y+z}(0,y,z)
&
P &= \frac 1{y+z}((1-x)^2,xy,xz)
\\
B' &= (x:0:z)=\frac 1{z+x}(x,0,z)
&
Q &= \frac 1{x+z}(xy,(1-y)^2,yz)
\\
C' &= (x:y:0)=\frac 1{x+y}(x,y,0)
&
R &= \frac 1{x+y}(xz,yz,(1-z)^2)
\end{aligned}
$$
The formula for the area $[S_1S_2S_3]$ of a triangle $\Delta S_1S_2S_3$ with vertices having barycentric coordinates $S_k(x_k,y_k,z_k)$ as a part of the area $[ABC]$ of $\Delta ABC$:
$$
\frac{[S_1S_2S_3]}{[ABC]}=
\begin{vmatrix}
x_1 & y_1 & z_1\\
x_2 & y_2 & z_2\\
x_3 & y_3 & z_3
\end{vmatrix}
\ .
$$ 
In our case we compute:
$$
\begin{aligned}
\frac{[A'B'C']}{[ABC]}
&=
\frac1{(y+z)(z+x)(x+y)}
\begin{vmatrix}
0&y&z\\
x&0&z\\
x&y&0
\end{vmatrix}
=
\\
&=
\frac
{2xyz}{(y+z)(z+x)(x+y)}\ ,
\\[3mm]
%
\frac{[PQR]}{[ABC]}
&=
\frac1{(y+z)(z+x)(x+y)}
\begin{vmatrix}
(1-x)^2 & xy & xz\\
xy & (1-y)^2 & yz\\
xz & yz & (1-z)^2
\end{vmatrix}
\\
&=
\frac
{2xyz}{(y+z)(z+x)(x+y)}\ .
\end{aligned}
$$
The first determinant is simple. For the second one we may force the factor $x$ in the first row, and in the first column, the same applies for $y,z$ and compute rather
$$
x^2y^2z^2
\begin{vmatrix}
(1-x)^2/x^2 & 1 & 1\\
1 & (1-y)^2/y^2 & 1\\
1 & 1 & (1-z)^2/z^2
\end{vmatrix}
\ .
$$
$\square$

Second solution, direct computations, but we keep the road closer to geometric operations and intuition. 
We have first for instance 
$$\frac{[AC'B']}{[ABC]}=\frac{AC'}{AB}\cdot\frac{AB'}{AC}
=\frac{yz}{(x+y)(x+z)}
=\frac{yz(y+z)}{(x+y)(y+z)(z+x)}
\ .
$$
This allows a quick computation of the ratio $[A'B'C']:[ABC]$ by expressing the area $[A'B'C']$ as $[ABC]-\sum [AB'C']$, and we get
$$
\begin{aligned}
\frac{[A'B'C']}{[ABC]}
&=
1
-\sum_{\text{cyclic}}
\frac{yz(1-x)}{(1-x)(1-y)(1-z)}
\\
&=
\frac{2xyz-x-y-z+1}{(1-x)(1-y)(1-z)}
=
\frac{2xyz}{(1-x)(1-y)(1-z)}
\end{aligned}
\ ,
$$
as in the first solution.
Let us compute now the contribution of $[OQR]:[ABC]$. We have, using signed areas this time:
$$
\frac{[OQR]}{[ABC]}
=
\frac{[OQR]}{[OBC]}
\frac{[OBC]}{[ABC]}
=
\frac{OQ}{OB}\cdot
\frac{OR}{OC}\cdot
\frac{OA'}{AA'}
=
\frac{1-2y}{1-y}\cdot
\frac{1-2z}{1-z}\cdot
\frac{x}{1}\ .
$$
Adding the three signed areas (since $1-2x$, $1-2y$, $1-2z$ may become negative), we obtain
$$
\begin{aligned}
\frac{[PQR]}{[ABC]}
&=
\sum_{\text{cyclic}}
\frac{[OQR]}{[ABC]}
=
\sum_{\text{cyclic}}
\frac{x(1-x)(1-2y)(1-2z)}{(1-x)(1-y)(1-z)}
\\
&=
\sum_{\text{cyclic}}
\frac{x(1-x)((1-y)-y)((1-z)-z)}{(1-x)(1-y)(1-z)}
\\
&=
\sum x-\sum\frac{xy}{1-y}-\sum\frac{xz}{1-z}+\sum\frac{xyz}{(1-y)(1-z)}
\\
&=
1-\sum\underbrace{\frac{(x+z)y}{1-y}}_{=y}
+ \frac{xyz}{(1-x)(1-y)(1-z)}\sum(1-x)
\\
&=
1-1+\frac{xyz}{(1-x)(1-y)(1-z)}\cdot 2
\ ,
\end{aligned}
$$
and computations lead to the same result.
$\square$
