Find length of transverse common tangent of two circles touching externally, intercepted by a third circle touching both circles externally. Two circles of radii $4$ and $16$ units, respectively touch each other externally. Prove that the length of the chord intercepted on the transverse common tangent to the to circles, by a circle of diameter $D$  units which touches the given circle externally, is $\frac{4D}{5}$


Here's an argument that could probably be streamlined a bit ...

Restating the problem:

Let $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ ---of radii $a$, $b$, $c$--- be pairwise externally tangent, with points of tangency $D$, $E$, $F$ as shown in the image. Let the line tangent to $\bigcirc A$ and $\bigcirc B$ at $F$ meet $\bigcirc C$ in chord $\overline{PQ}$ with midpoint $M$; define $p := |\overline{PM}| = |\overline{QM}|$. Then
$$p = \frac{2c\sqrt{ab}}{a+b} \tag{$\star$}$$


Extend $\overline{FE}$ and $\overline{FD}$ to meet $\bigcirc C$ again at $A^\prime$ and $B^\prime$. Note that $\triangle AEF$ and $\triangle CEA^\prime$ are isosceles, with congruent base angles at $E$ and therefore also at $F$ and $A^\prime$; thus, $\overline{A^\prime C}\parallel\overline{AF}$. Likewise, $\overline{B^\prime C}\parallel\overline{BF}$, making $A^\prime$, $B^\prime$, $C$ collinear on a diameter of $\bigcirc C$.
By Thales' Theorem, $\angle A^\prime PB^\prime$ is a right angle; because $\overline{PM}\perp\overline{A^\prime B^\prime}$, we have the classical construction of a geometric mean:
$$|\overline{PM}|^2 = |\overline{A^\prime M}||\overline{B^\prime M}| \tag{1}$$
(This is why I chose $p$ to name the half-chord, instead of the full length of $\overline{PQ}$.) Now our task reduces to a matter of rewriting the right-hand side.
Observe that the perpendiculars at $D$, $E$, $F$ necessarily concur at a point equidistant from them; that is, they meet at center $I$ of the incircle (of radius $r$) of $\triangle ABC$. Of course, $\overline{AI}$ and $\overline{BI}$ are angle bisectors.
Since $\triangle A^\prime M F\sim \triangle IFA$ and $\triangle B^\prime MF\sim \triangle IFB$, we have
$$\begin{align}
\frac{|\overline{A^\prime M}|}{|\overline{MF}|} = \frac{r}{a} \quad\text{and}\quad 
\frac{|\overline{B^\prime M}|}{|\overline{MF}|} = \frac{r}{b} &\quad\implies\quad
\frac{|\overline{B^\prime M}|}{|\overline{A^\prime M|}} = \frac{a}{b} \tag{2} \\
&\quad\implies\quad
|\overline{A^\prime M}| = 2c\;\frac{b}{a+b} \quad\text{and}\quad
|\overline{B^\prime M}| = 2c\;\frac{a}{a+b} 
\end{align}$$
(The above improves on my first pass at this proof. (See the revision history.) There may yet be a better argument that incorporates this proportionality relation:
$$\frac{|\overline{A^\prime M}|}{b} = \frac{|\overline{A^\prime B^\prime}|}{|\overline{AB}|} = \frac{|\overline{B^\prime M}|}{a} \tag{2.5}$$
Be that as it may ...) Thus, $(1)$ becomes
$$p^2 = \frac{4abc^2}{(a+b)^2} \qquad\to\qquad p = \frac{2c\sqrt{ab}}{a+b} \tag{3}$$
as claimed. $\square$
For the original question, we take $a = 4$, $b = 16$, $c = D/2$, so that the length of the chord is 
$$2p = \frac{4 \cdot D/2 \cdot \sqrt{4\cdot 16}}{4+16} = \frac{4D}{5}$$
