Problem source

The vertices $A$, $B$, $C$ and $D$ of a square have coordinates $(0,0)$, $(a,0)$, $(a,a)$ and $(0,a)$, respectively.
The points $P$ and $Q$ have coordinates $(an,0)$ and $(0,am)$ respectively, where $0 < m < n < 1$.  
The line $CP$ produced meets $DA$ produced at $R$ and the line $CQ$ produced meets $BA$ produced at $S$.  The line $PQ$ produced meets the line $RS$ produced at $T$.
Show that $TA$ is perpendicular to $AC$.
Explain how, given a square of area $a^2$, a square of area $2a^2$ may be constructed using only a straight-edge.
 
[\textbf{Note}: a straight-edge is a ruler with no markings on it;
no measurements (and no use of  compasses) are allowed in the construction.]

Solution source

\begin{center}
\begin{tikzpicture}

\def\a{3};
\def\n{.3};
\def\m{.45};

\coordinate (A) at (0,0);
\coordinate (B) at (\a,0);
\coordinate (C) at (\a,\a);
\coordinate (D) at (0,\a);

\coordinate (P) at ({\a*\n}, 0);
\coordinate (Q) at (0, {\a*\m});

\coordinate (R) at (0, {-\a*\n/(1-\n)});
\coordinate (S) at ({-\a*\m/(1-\m)}, 0);
\coordinate (T) at ({\a*\m*\n/(\m-\n)}, {-\a*\m*\n/(\m-\n)});

\filldraw (A) circle (1.5pt) node[below left] {$A$};
\filldraw (B) circle (1.5pt) node[below right] {$B$};
\filldraw (C) circle (1.5pt) node[above right] {$C$};
\filldraw (D) circle (1.5pt) node[above left] {$D$};
\filldraw (P) circle (1.5pt) node[below] {$P$};
\filldraw (Q) circle (1.5pt) node[left] {$Q$};
\filldraw (R) circle (1.5pt) node[left] {$R$};
\filldraw (S) circle (1.5pt) node[below] {$S$};
\filldraw (T) circle (1.5pt) node[below left] {$T$};


\draw (A) -- (B) -- (C) -- (D) -- cycle;

\draw[dashed] (A) -- ($(A)!-0.6!(D)$);
\draw[dashed] (A) -- ($(A)!-0.9!(B)$);
\draw[dashed] (Q) -- ($(P)!-2.5!(Q)$);
\draw[dashed] (S) -- ($(R)!-1.5!(S)$);
\draw[dashed] (C) -- (R);
\draw[dashed] (C) -- (S);

\end{tikzpicture}
\end{center}

Note that $CP$ has equation $\frac{y-0}{x-an} = \frac{a-0}{a-an} = \frac{1}{1-n} \Rightarrow y = \frac{x-an}{1-n}$
Therefore $R = (0, -\frac{an}{1-n})$

Note that $CQ$ has equation $\frac{y-am}{x} = \frac{a-am}{a} = 1-m \Rightarrow y = (1-m)x + am$
Therefore $S = (-\frac{am}{1-m}, 0)$

$PQ$ has equation $\frac{y}{x-an} = \frac{am-0}{0-an} \Rightarrow y = -\frac{m}{n}x +am$

$SR$ has equation $\frac{y}{x+\frac{am}{1-m}} = \frac{-\frac{an}{1-n}}{\frac{am}{1-m}} = -\frac{n(1-m)}{m(1-n)} \Rightarrow y =-\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n}$

So $PQ \cap SR$ has 
\begin{align*}
&& -\frac{m}{n}x +am &= -\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n} \\
&& x \left (\frac{n(1-m)}{m(1-n)} - \frac{m}{n} \right) &= -am - \frac{an}{1-n} \\
\Rightarrow && x \left ( \frac{n^2(1-m)-m^2(1-n)}{nm(1-n)} \right) &= -\frac{a(m(1-n)+n)}{1-n} \\
\Rightarrow && x \left ( \frac{(m-n)(mn-n-m)}{mn(1-n)} \right) &= \frac{a(mn-m-n)}{1-n} \\
\Rightarrow && x &= \frac{amn}{m-n} \\
&& y &= -\frac{amn}{m-n}
\end{align*}

Therefore clearly $TA$ is perpendicular to $AC$ since they are the lines $y = -x$ and $y = x$

Given this method we can construct the perpendicular to the diagonal through the vertex. Doing this at $A$ we can construct $C'$ the reflection of $C$ in $AB$.

We can do the same to find the reflection of $A$ and so we have a square with sidelengths $\sqrt{2}a$ and hence area $2a^2$

\begin{center}
\begin{tikzpicture}

\def\a{3};
\def\n{.3};
\def\m{.45};

\coordinate (A) at (0,0);
\coordinate (Aa) at ({\a*2},0);
\coordinate (B) at (\a,0);
\coordinate (C) at (\a,\a);
\coordinate (Ca) at (\a,-\a);
\coordinate (D) at (0,\a);

\filldraw (A) circle (1.5pt) node[below left] {$A$};
\filldraw (B) circle (1.5pt) node[below right] {$B$};
\filldraw (C) circle (1.5pt) node[above right] {$C$};
\filldraw (D) circle (1.5pt) node[above left] {$D$};
\filldraw (Ca) circle (1.5pt) node[below right] {$C'$};
\filldraw (Aa) circle (1.5pt) node[above right] {$A'$};

\draw (A) -- (B) -- (C) -- (D) -- cycle;

\draw[dashed] (Aa) -- (C) -- (A) -- (Ca) -- cycle;
\draw[dashed] (B) -- ($(B)!-1.4!(C)$);
\draw[dashed] (B) -- ($(B)!-1.4!(A)$);
\end{tikzpicture}
\end{center}