It's not very easy (only one person was able to solve it completely in Slovakia), but maybe for you it will be easy.

ondrob wrote: Let $ABCD$ be a trapezoid such that $AB$ is parallel to $CD$, and let $E$ be the midpoint of its side $BC$. Suppose we can inscribe a circle into the quadrilateral $ABED$, and that we can inscribe a circle into the quadrilateral $AECD$. Denote $|AB|=a$, $|BC|=b$, $|CD|=c$, $|DA|=d$. Prove that \[a+c=\frac{b}{3}+d;\] \[\frac{1}{a}+\frac{1}{c}=\frac{3}{b}\] Denote $AE = x, DE = y$. Since the quadrilaterals $ABED, AECD$ are both tangential, the sums of their opposite sides are equal: $a+y = \frac{b}{ }2+d = x+c$ First, assume that the trapezoid $ABCD$ is a rectangle. Then $a = c, b = d, x = y$. Let $k = \frac{b}{ }a$ be the aspect ratio of the rectangle: $b = d = ka,\ \ x = y = \sqrt{a^{2}+\frac{b^{2}}{4}}= a \sqrt{1+\frac{k^{2}}{4}}$ $a+a \sqrt{1+\frac{k^{2}}{4}}= \frac{ka}{2}+ka$ $1+\frac{k^{2}}{4}= \left(\frac{3k}{2}-1\right)^{2}= \frac{9k^{2}}{4}-3k+1$ $2k^{2}-3k = 0,\ \ k = \frac{3}{ }2$ . or . $0$ The zero root can be ignored, it is not a root of the original equation before we squared it. If the aspect ratio is $k = \frac{3}{ }2$, then $b = d = \frac{3a}{2}$, $x = y = a \sqrt{1+\frac{9}{16}}= \frac{5a}{4}$ and $\frac{b}{ }3+d = \frac{a}{ }2+\frac{3a}{2}= 2a = a+c$ $\frac{3}{ }b = \frac{2}{ }a = \frac{1}{ }a+\frac{1}{ }c$ It is clear that the trapezoid $ABCD$ cannot be any other non-rectangular parallelogram, because then $a = c, b = d, x \neq y$ and the quadrilaterals $ABED, AECD$ cannot be simultaneously tangential. Assume that $ABCD$ is a general trapezoid. Let $(I_{1}), (E_{2})$ be the incircles of the quadrilaterals $ABED, AECD$ (the purpose of this notation will become clear later on). Let $P_{1}, Q_{1}, R_{1}, S_{1}$ be the tangency points of the circle $(I_{1})$ with the sides $AB, BE, ED, DA$ and $P_{2}, Q_{2}, R_{2}, S_{2}$ the tangency points of the circle $(E_{2})$ with the sides $AE, EC, CD, DA$, respectively. Let $B', C'$ be the intersections of the lines $DE, AE$ with the parallel lines $AB, CD$. The quadrilateral $AB'C'D$ is a parallelogram with the sides $AB' = C'D = a+c,\ B'C' = DA = d$ and the diagonal intersection $E \equiv AC' \cap B'D$. This is because the $E$ is the midpoint of the segment $BC$. Consequently, the triangles $\triangle ABE \cong \triangle C'CE$ and the triangles $\triangle DCE \cong \triangle B'BE$ are congruent by ASA and the diagonals $AC', B'D$ cut each other in half. The circles $(I_{1}), (E_{2})$ are the incircles of the triangles $\triangle AB'D, \triangle AC'D$. Let $s_{1}= \frac{a+c+d+2y}{2}$, $s_{2}= \frac{a+c+d+2x}{2}$ be the semiperimeters of these 2 triangles. The tangent lengths $ER_{1}, EP_{2}$ to the circles $(I_{1}), (E_{2})$ from the diagonal intersection $E$ are equal to $ER_{1}= DE-DR_{1}= y-(s_{1}-AB') =$ $= y-\left[\frac{a+c+d+2y}{2}-(a+c)\right] = \frac{a+c-d}{2}$ $EP_{2}= AE-AP_{2}= x-(s_{2}-C'D) =$ $= x-\left[\frac{a+c+d+2x}{2}-(a+c)\right] = \frac{a+c-d}{2}$ Since the tangent lengths $ER_{1}= EP_{2}$ are equal, the point $E$ is on the radical axis of the circles $(I_{1}), (E_{2})$. This means that the tangent lengths $EQ_{1}= EQ_{2}= ER_{1}= EP_{2}= \frac{a+c-d}{2}$ form the point $E$ to the circles $(I_{1}), (E_{2})$ are all equal. The segment lengths $Q_{1}Q_{2}= S_{1}S_{2}= 2EQ_{1}= a+c-d$ of their common external tangents $DA, BC$ are also equal. Hence, the segment $d = DA$ is equal to $d = DA = AS_{2}+DS_{1}-S_{1}S_{2}= AP_{2}+DR_{1}-S_{1}S_{2}=$ $= AE-EP_{2}+DE-ER_{1}-S_{1}S_{2}= x+y-2(a+c-d)$ If we add the original 2 equations for the equal sums of the opposite sides of the tangential quadrilaterals $ABED, AECD$, we obtain $a+y+c+x = 2d+b,\ \ x+y = 2d+b-(a+c)$ Substituting this to the above equation leads to $d = 2d+b-(a+c)-2(a+c-d),\ \ 3(a+c) = b+3d,\ \ a+c = \frac{b}{ }3+d$ . . It follows that the segment lengths $Q_{1}Q_{2}= S_{1}S_{2}$ of the common external tangents of the circles $(I_{1}), (E_{2})$ are equal to $Q_{1}Q_{2}= S_{1}S_{2}= a+c-d = \frac{b}{ }3$ Since $E$ is the midpoint of the segment $BC$, the tangent lengths $BP_{1}= BQ_{1}$ to the circle $(I_{1})$ from the vertex $B$ are equal to the tangent lengths $CQ_{2}= CR_{2}$ from the vertex $C$ to the circle $(E_{2})$, all of them being equal to $BP_{1}= BQ_{1}= CQ_{2}= CR_{2}= BE-EQ_{1}= \frac{b}{ }2-\frac{b}{ }6 = \frac{b}{ }3 = Q_{1}Q_{2}$ Let $H$ be the intersection of the lines $BC, DA$, the common external tangents of the circles $(I_{1}), (E_{2})$. This is the external homothety center of these 2 circles and also the homothety center of the centrally similar triangles $\triangle ABH \sim \triangle DCH$. Let $\eta = \frac{HA}{HD}= \frac{HB}{HC}= \frac{AB}{DC}= \frac{a}{ }c$ be the homothety coefficient of these 2 triangles. The circle $(I_{1})$ is the incircle of the triangle $\triangle ABH$ and the circle $(E_{2})$ is the excircle of the triangle $\triangle DCH$ against the vertex $H$. Let $(E_{1})$ be the excircle of the triangle $\triangle ABH$ against the vertex $H$ tangent to the side $AB$ at a point $T_{1}$ and $(I_{2})$ the incircle of the triangle $\triangle DCH$ tangent to the side $DC$ at a point $T_{2}$. The circles $(I_{1}) \sim (I_{2})$ and the circles $(E_{1}) \sim (E_{2})$ are centrally similar with the same homothety center $H$ and coefficient $\eta$ as the triangles $\triangle ABH \sim \triangle DCH$. Let $\sigma_{1}, \sigma_{2}$ be the semiperimeters of these 2 triangles, $\sigma_{1}= \eta \sigma_{2}$. The tangent length $BP_{1}$ to the incircle $(I_{1})$ from the vertex $B$ is equal to the tangent length $AT_{1}$ to the excircle $(E_{1})$ from the vertex $A$: $BP_{1}= \sigma_{1}-HA = AT_{1}$ Similarly, the tangent length $CT_{2}$ to the incircle $(I_{2})$ from the vertex $C$ is equal to the tangent length $DR_{2}$ to the excircle $(E_{2})$ from the vertex $D$: $CT_{2}= \sigma_{2}-HC = DR_{2}$ From the central similarity of the triangles $\triangle ABH \sim \triangle DCH$ with their incircles $(I_{1}), (I_{2})$ and excircles $(E_{1}), (E_{2})$, we have $\frac{AP_{1}}{BP_{1}}= \frac{AP_{1}}{CR_{2}}= \frac{AP_{1}}{DT_{2}}= \eta$ $\frac{CR_{2}}{DR_{2}}= \frac{DT_{2}}{CT_{2}}= \frac{AP_{1}}{BP_{1}}= \eta$ Hence, the tangency points $P_{1}, R_{2}$ divide the sides $AB, CD$ in the same ratio equal to the homothety coefficient $\eta$. As a result, $a = AB = AP_{1}+BP_{1}= \eta BP_{1}+BP_{1}= \left(\frac{a}{ }c+1\right) \frac{b}{ }3$ $\frac{ac}{a+c}= \frac{b}{ }3,\ \ \frac{1}{ }a+\frac{1}{ }c = \frac{3}{ }b$ . . Ahoj, Yetti

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Yeah, but it looks quite long I will describe as I solved it (but I'm a bit lazy to write it all down, maybe later ) We know, when we denote sides of $ABCD$ like $a,b,c,d$ to compute lengths of $DE,EB',BB'$ in terms of $a,b,c,d$ easily (using tangentiality). So we can easily compute semiperimeters of $\triangle AB'D,\triangle BB'E$ in terms of $a,b,c,d$ (denote them $p_1,p_2$) and also denote $|BE|$ as $x$. Also denote areas of $\triangle AB'D,\triangle BB'E$ as $P_1,P_2$. Now using that circle inscribed in $ABED$ is incircle of $\triangle AB'D$ and excircle of $\triangle BB'E$ we get: $\frac{P_1}{p_1}=\frac{P_2}{p_2-x} $ We can rewrite it as: $\frac{P_1}{P_2}=\frac{p_2-x}{p_1}$ But $\frac{P_1}{P_2}$ we can easily compute as: $\frac{P_1}{P_2}=\frac{AB'}{BB'}\frac{DB'}{EB'}=\frac{2(a+c)}{c}$ Using this in $\frac{P_1}{P_2}=\frac{p_2-x}{p_1}$ we get: $2(a+c)p_1=c(p_2-x)$ But we can write it in terms of $a,b,c,d$ with $b,d$ being just in linear form! Analogicaly we can get another linear equation of $b,d$ by using that the second circle is inscribed and excribed at the same time. From them we can express $b,d$ (just solve system of 2 linear equations) and we get: $b=\frac{3ac}{a+c}\qquad d=\frac{a^2+ac+c^2}{a+c}$ Putting this in the two equations we get true statements. If someone can't find out how exactly I did it or something, just tell, I'll find time to write it completely.