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<h1>An equilateral triangle inscribed in a rectangle</h1>
<p>
The following solution to the
<a href="http://userpages.umbc.edu/~rostamia/Geometry/inscribed-equilateral.html">inscribed triangle puzzle</a>
is due to <b>Peter&nbsp;Renz</b>
who communicated it to me on December&nbsp;2016.
<p>
For the sake of completeness, let's begin with the statement of the puzzle.
The figure below depicts a equilateral triangle $AEF$ inscribed in a rectangle
in such a way that the two share a vertex. We wish to show that the
area of the pink triangle ($ECF$) is the sum of the areas of the other two
colored triangles.
<div class="centered">
<img src="./inscribed-equilateral-solution/frame000.png" width=500
height=417 alt="[frame000.png]">
</div>
<p>
As noted in the referring page (see the link above), a solution with
the aid of trigonometry is quite straightforward. The purpose of this
page is to present a solution in the style of <i>Euclid</i>, without appeal
to trigonometry.
<p>
The following animation encapsulates Peter&nbsp;Renz's solution in its
entirety.
<div class="centered">
<img src="./inscribed-equilateral-solution/waterwheel.gif" width=500
height=417 alt="[animation.gif]">
</div>
The animation should be self-explanatory if you stare at it long enough.
Nevertheless, I will now proceed to point out the reasoning through
several still images extracted from that animation.
<ol>
<li>
Construct a circle on the diameter $FE$, and then subdivide its boundary
into six 60-degree wedges starting at the vertex $C$. Mark the division
points $C$, $P$, $Q'$, $C'$, $P'$, and $Q$, as shown.
<div class="centered">
<img src="./inscribed-equilateral-solution/frame007.png" width=500
height=417 alt="[frame007.png]">
</div>
<li>
Rotate the triangle $FDA$ about $F$ by 60 degrees to bring the edge $FA$ to
coincide with $FE$.
<div class="centered">
<img src="./inscribed-equilateral-solution/frame073.png" width=500
height=417 alt="[frame073.png]">
</div>
The key observation is that the rotation moves the vertex $D$ into $P'$.
<li>
Further rotate the triangle about the circle's center by 180 degrees to
place it in the $FPE$ position.
<div class="centered">
<img src="./inscribed-equilateral-solution/frame111.png" width=500
height=417 alt="[frame111.png]">
</div>
<li>
Rotate the triangle $EBA$ about $E$ by 60 degrees to bring the edge $EA$ to
coincide with $EF$. Then the vertex $B$ will move onto $Q'$ for reasons
similar to those explained above.
<div class="centered">
<img src="./inscribed-equilateral-solution/frame177.png" width=500
height=417 alt="[frame177.png]">
</div>
<li>
The two red line segments in the figure below are parallel and of equal lengths
by virtue of being the side and the &ldquo;radius&rdquo; of the
regular hexagon (not shown) inscribed in the circle.
<div class="centered">
<img src="./inscribed-equilateral-solution/frame180.png" width=500
height=417 alt="[frame180.png]">
</div>
The altitudes of the three triangles, dropped from the vertices $C$,
$P$, and $Q'$ are shown in dashed lines. It should be clear that the
altitude dropped from $C$ is equal in length to the sum of those of
the other two altitudes. Since the three triangles share a common base,
the area of one is the sum of the areas of the other two. Q.E.D.
</ol>
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