Pessimist Singularitarian
#4
December 7th, 2016,
00:47
Okay, let's label the unknown angles:
\begin{tikzpicture}
\draw[blue,thick] (0,0) -- (12,0);
\draw[blue,thick] (12,0) -- (8,4);
\draw[blue,thick] (8,4) -- (0,0);
\draw[blue,thick] (8,4) -- (6,0);
\node[left=5pt of {(0,0)}] {\large P};
\node[above=5pt of {(8,4)}] {\large Q};
\node[below=5pt of {(6,0)}] {\large R};
\node[right=5pt of {(12,0)}] {\large S};
\node[right=15pt of {(0,0)},yshift=5pt] {\large $x$};
\node[right=5pt of {(6,0)},yshift=7pt] {\large $2x$};
\node[left=5pt of {(8,4)},yshift=-13pt] {\large $\alpha$};
\node[below=3pt of {(8,4)}] {\large $\beta$};
\node[left=10pt of {(12,0)},yshift=7pt] {\large $\delta$};
\node[above=0pt of {(6,0)},xshift=-3pt] {\large $\gamma$};
\end{tikzpicture}
Now, we know the following:
$ \displaystyle 2x+\gamma=180^{\circ}$
$ \displaystyle x+\alpha+\gamma=180^{\circ}$
$ \displaystyle 2x+\beta+\delta=180^{\circ}$
$ \displaystyle x+\alpha+\beta+\delta=180^{\circ}$
We need to show that:
$ \displaystyle x=\alpha$
Look at the first two equations above, and we see that:
$ \displaystyle 180^{\circ}-\gamma=2x=x+\alpha$
What does this imply?