Tangential quadrilateral and cyclic quadrilateral Let $ABCD$ be a convex quadrilateral on whose sides $AB, BC, CD$ and $DA$ points $E, F, G$ and $H$ are located in such a way that they divide the corresponding side in relation to the lengths of the adjacent sides, i.e.  
$$\frac {AE}{EB}=\frac {DA}{BC}, \frac {BF}{FC}=\frac {AB}{CD}, \frac {CG}{GD}=\frac {BC}{DA}, \frac {DH}{HA}=\frac {CD}{AB}.$$
Show that under these conditions the following always holds true: 
$ABCD$ is a tangential quadrilateral if and only if $EFGH$ is a cyclic quadrilateral. 
My second attempt:
Rearranging the ratios and with some manipulation I get 
$$\frac{AE}{HA}=\frac{EB}{BF}=\frac{CG}{FC}=\frac{DH}{GD}=\frac{AB+CD}{BC+DA}.\tag{1}$$
If $AB + CD = BC + DA$ then $ABCD$ is tangential. The ratios in (1) are equal to one and the sides of $ABCD$ are tangents to the incircle. With $E, F, G$ and $H$ lying on the sides of $ABCD$ the incircle becomes a cirumcircle of $EFGH$. Thus $EFGH$ is cyclic.
Now the other direction: If $EFGH$ is cyclic, then its opposing inner angles add up to 180°. 

The blue angles then also add up to 180° as do the red angles (but note they are not equal). This is only possible if the ratios in above equation equal one but then $ABCD$ must be tangential (to be seen if on assumes that the ratios do not equal one).
 A: Hint: Draw a line through the midpoints, $M_1$ and $M_2$, of the diagonals of the quadrilateral $ABCD$. On this line, there exits a point, $O$, the center of the inscribed circle of this quadrilateral:

Since each line is tangent to this circle, then the following must be true:
$$|AE|=|AH|, |BE|=|BF|, |CF|=|CG|, |DG|=|DH|$$
$$\implies |AB| + |CD| = |BC| + |AD|$$
The center of each inscribed ellipse will be on the line $|M_1M_2|$:

A: Before we begin, let's clarify what seems to be a common point of confusion about the problem: As the diagram shows, the incircle of $\square ABCD$ is not (necessarily) the same as the circumcircle of $\square EFGH$. (For convex $\square ABCD$, the circles are concentric.)

Now, we'll define
$$a\phantom{^\prime} := |AB| \qquad b\phantom{^\prime} := |BC| \qquad c\phantom{^\prime} := |CD| \qquad d\phantom{^\prime} := |DA|$$
$$e\phantom{^\prime} := |EB| \qquad f\phantom{^\prime} := |FC| \qquad g\phantom{^\prime} := |GD| \qquad h\phantom{^\prime} := |HA|$$
$$e^\prime := |AE| \qquad f^\prime := |BF| \qquad g^\prime := |CG| \qquad h^\prime := |DH|$$
From the given proportions, we deduce via simple algebra:
$$\begin{align}e\phantom{^\prime} &= \frac{a b}{b + d} \qquad f\phantom{^\prime} = \frac{b c}{ a + c } 
\qquad g\phantom{^\prime} = \frac{c d}{ b + d } 
\qquad h\phantom{^\prime} = \frac{d a}{ a + c } \\[4pt]
e^\prime &= \frac{da}{ b + d }
\qquad f^\prime = \frac{ab}{ a + c }
\qquad g^\prime = \frac{bc}{ b + d }
\qquad h^\prime = \frac{cd}{ a + c }
\end{align} \tag{1}$$
This gives us one direction of the result fairly quickly.
$$\begin{align} \square ABCD \;\text{tangential} &\implies
a+c=b+d \\
&\implies e = f^\prime, f = g^\prime, g = h^\prime, h = e^\prime \\
&\implies \angle AEH = \angle AHE, \;\text{etc} \\
&\implies \angle AEH = 90^\circ - \frac12 A, \;\text{etc} \\
&\implies
\angle FEH + \angle FGH = \left(180^\circ - \frac12(A+B)\right) + \left(180^\circ - \frac12(C+D)\right) \\
&\implies \angle FEH + \angle FGH = 360^\circ - \frac12(A+B+C+D) = 180^\circ \\
&\implies \square EFGH \;\text{cyclic} \quad\text{(QED)}
\end{align}$$
(Note: When $\square ABCD$ is convex, the fact that $\triangle AEH$, $\triangle BFE$, $\triangle CGF$, $\triangle DHG$ are isosceles tells us that $\square EFGH$'s circumcenter lies on the bisectors of $\angle A$, $\angle B$, $\angle C$, $\angle D$. Therefore, the circumcenter coincides with $\square ABCD$'s incenter. The circles are concentric, as claimed above.)
The converse seems a bit trickier. A cyclic $\square EFGH$ tells us that $$\angle AEH + \angle BEF + \angle CGF + \angle DGH = 180^\circ = \angle AHE + \angle BFE + \angle CFH + \angle DHG$$
and the relations $(1)$ and the Law of Sines tell us, for instance:
$$\frac{\sin\angle AEH}{\sin\angle AHE} = \frac{\sin\angle BFE}{\sin\angle BEF} = \frac{\sin\angle CGF}{\sin\angle CFG} = \frac{\sin\angle DHG}{\sin\angle DGH} = \frac{a+c}{b+d}$$
An easy approach to showing that the ratio is $1$ has eluded me so far.
But a complicated approach? Oh, I have one of those ...

Some intense coordinate-based symbol-crunching leads from $(1)$ to these relations:
$$|EG|^2 = \frac{bd}{(b+d)^2} \left( r^2 - (a+c)^2 \right) \qquad
|FH|^2 = \frac{ac}{(a+c)^2} \left( r^2 - (b+d)^2 \right) \tag{2}$$
where $r^2 := |AC|^2 + |BD|^2 + 2(a c + b d)$. Further, if $\overline{EG}$ and $\overline{FH}$ meet at $K$, then
$$|EK| = \frac{a}{a+c}|EG| \qquad |GK| = \frac{c}{a+c}|EG| \qquad
|FK| = \frac{b}{b+d}|FH| \qquad |HK| = \frac{d}{b+d}|FH|$$
Consequently,
$$|EK||KG|-|FK||KH| = \frac{abcd}{(a+c)^2(b+d)^2}\left((b+d)^2-(a+c)^2\right)$$
which is to say

$$|EK| |KG| = |FK||KH|  \quad\iff\quad a+c=b+d \tag{$\star$}$$

We know that the right-hand component of $(\star)$ is equivalent to "$\square ABCD$ is tangential"; an aspect of the power of a point theorem implies that the left-hand component is equivalent to "$\square EFGH$ is cyclic". Thus, $(\star)$ demonstrates both directions of the desired result at once! $\square$
I'd be pretty happy with this approach, if not for the difficulty of deriving $(2)$. Given that this exercise is targeted for students of "basic geometry", I suspect that I must be missing something.  
