Interpreting the geometric/harmonic/arithmetic means of two numbers as lengths in a trapezoid Let $a$ and $b$ be any two (distinct, positive) numbers. We can interpret the arithmetic mean (AM) and the harmonic mean (HM) of these numbers as line segments parallel to the bases of a trapezoid of lengths $a$ and $b$: specifically, the AM segment passes through the midpoints of the legs; the HM segment passes through the intersection of the diagonals.

How does one show, geometrically, that the corresponding parallel line segment representing the geometric mean (GM) of $a$ and $b$ lies between the line segments representing the other two means?
Moreover, how do we show that the HM segment is closer to the trapezoid's shorter base, while the AM segment is closer to the longer base? That is, how can we use this trapezoid interpretation to establish the inequality "$\text{HM} < \text{GM} <\text{AM}$"?

 A: I'll assume that the lengths $a$ and $b$ are unequal, for otherwise
the three means are all the same.
For convenience in describing the constructions, assume without loss of
generality that $a < b$.
The difficulty with this problem in my opinion is that while the
constructions of the arithmetic and harmonic means have simple constructions
that directly use the form of the trapezoid itself (namely, the midpoints
of the lateral sides and the intersection of the diagonals, respectively),
the geometric mean does not.
There is a relatively simple construction of the geometric mean of the
two bases of a trapezoid on the page linked here.
In first attempting to answer this question I followed a very similar construction (although with different labels for the points), 
but I choose now to use a variation of the construction that makes use of some steps that we would otherwise have to add later in order to show that the geometric mean is between the other two means.
My construction, however, will still use the same basic idea as the other construction, which is to first find a segment
of length $\sqrt{ab}$ (the geometric mean) somewhere on the plane,
then find an equal segment that is parallel to $\overline{AB}$ and $\overline{CD}$ and has endpoints on $\overline{BC}$ and $\overline{AD}$.
In order to find a segment of length $\sqrt{ab}$, the first step is to
construct a segment of length $a$ adjacent to segment $\overline{CD}$.
One way to do this is to construct the midpoint $M$ of side $\overline{AD}$ and extend the lines $\overleftrightarrow {BM}$ and $\overleftrightarrow {CD}$ until they intersect at $E$.
The point $D$ now divides the segment $\overline{CE}$ into segments
of lengths $CD=b$ and $DE=a$.
We construct a semicircle using $\overline{CE}$ as its diameter, and
extend a line perpendicular to $\overline{CE}$ through $D$ until it
intersects the semicircle at $F$. Then $DF = \sqrt{ab}$.
We have then already constructed the geometric mean of $a$ and $b$,
although not in the location in the plane where we ultimately want
to construct it.
To construct an equal-length segment at the desired location,
we can use a compass to construct a point $G$ on segment $\overline{CD}$ such that $DG = DF$. (We know there is such a point between $C$ and $D$ because $a < b$ and therefore $DF = \sqrt{ab} < b = CD$.)
Next, we find the intersection of segment $\overline{AG}$ with diagonal $\overline{BD}$ at $K$.
Finally, we construct the line through $K$ parallel to $\overline{AB}$ and $\overline{CD}$ and find its points of intersection $P$ and $Q$ with the sides of the trapezoid $\overline{AD}$ and $\overline{BC}$, respectively.

Now triangles $\triangle ABK$ and $\triangle GDK$ are similar, and
their bases $AB$ and $DG$ (hence also their corresponding altitudes,
and the altitudes of trapezoids $ABQP$ and $CDPQ$)
are in the ratio $a:\sqrt{ab}$.
Therefore $AB:PQ = PQ:CD = a:\sqrt{ab}$, and $PQ = \sqrt{ab}$; we have constructed the geometric mean of $a$ and $b$ at the desired location.
For the next step 
(showing that $\frac{2ab}{a+b} < \sqrt{ab} < \frac{a+b}{2}$),
we construct point $H$ on segment $\overline{CD}$ such that $DH = AB$.
(We can do this by drawing a compass arc from $E$ around center $D$ until
the arc intersects $\overline{CD}$ at $H$.)
Since $a < \sqrt{ab} < b$, we have
$DH < DG < CD$, and therefore $G$ is between $C$ and $H$.
Now let $J$ be the intersection of segment $\overline{AC}$ with diagonal $\overline{BD}$, and construct segment $\overline{MN}$ through $J$ with
endpoint $N$ on side $\overline{BC}$ of the trapezoid. Then $\overline{MN}$
is parallel to $\overline{AB}$ and $\overline{CD}$ (one reason is that $M$ and $J$ bisect $AD$ and $AH$), and $MN$ is the arithmetic mean of $a$ and $b$.
Next, let $L$ be the intersection of diagonal $overline{AC}$ with diagonal $overline{BD}$, and construct segment $\overline{RS}$ through $J$
parallel to $\overline{AB}$ and $\overline{CD}$ so that the endpoints
$R$ and $S$ are on the trapezoid's sides $\overline{AD}$ and $\overline{BC}$,
respectively. Then $RS$ is the harmonic mean of $a$ and $b$.
Now the points $J$, $K$, and $L$ all lie along the diagonal $\overline{BD}$,
and they are on the line segments from $A$ to
(respectively) $H$, $G$, and $C$.
But $H$, $G$, and $C$ occur in that order along $\overline{CD}$, and
therefore $J$, $K$, and $L$ also occur in that order along $\overline{BD}$,
and the segment $\overline{PQ}$ through $K$, 
representing the geometric mean, 
lies between the segments representing the harmonic and arithmetic means.
A: In a trapezoid with bases of lengths "a" and "b," the geometric mean, "G," is the length of the segment that is parallel to the bases and that also divides the trapezoid into two similar trapezoids.
Source: http://jwilson.coe.uga.edu/emt668/emat6680.2000/umberger/EMAT6690smu/Essay3smu/Essay3smu.html
So if you take that for granted, it must be between the two bases so that it can divide the trapezoid.
