$ ABCD$ is the convex quaddrilateral, $ (I), (J)$ are incircles of the $ \triangle DAB, \triangle BCD$ tangent to $ BD$ at $ U, V.$ The internal homothety center of $ (I), (J)$ is $ H \equiv IJ \cap BD.$ If $ \angle DAB = \angle BCD,$ either condition $ BU = DV$ or $ H \equiv IJ \cap BD \cap AC$ makes $ ABCD$ is a parallelogram and the proposition becomes trivial. Therefore, WLOG, let $ \angle DAB < \angle BCD$ and $ P \equiv AB \cap CD.$ Assume $ BU = DV.$ Then $ AB + BD - DA = CD + DB - BC$ or $ AB + BC = CD + DA,$ which means that the convex $ ABCD$ is tangential with an excircle $ (E)$ in the (smaller) angle $ \angle DAB.$ $ A$ is the external homothety center of $ (I), (E)$ and $ C$ the internal homothety center of $ (J), (E),$ so that the internal homothety center $ H$ of $ (I), (J)$ is on $ AC,$ therefore $ H \equiv IJ \cap BD \cap AC.$ Assume $ H \equiv IJ \cap BD \cap AC.$ Let $ (E)$ be the A-excircle of the $ \triangle DAP.$ The internal homothety center of $ (J), (E)$ is on their common internal tangent $ CD \equiv PD.$ $ A$ is the external homothety center of $ (I), (E)$ and $ H \in AC$ the internal homothety center of $ (I), (J)$ $ \Longrightarrow$ the internal homotherty center of $ (J), (E)$ is on $ AC$ $ \Longrightarrow$ it is identical with $ C \equiv CD \cap AC.$ But then the other tangent $ BC$ of $ (J)$ through $ C$ is also tangent of $ (E)$ and the convex $ ABCD$ is tangential with the excircle $ (E)$ in the angle $ \angle DAB.$ Consequently, $ AB + BC = CD + DA$ or $ AB + BD - DA = CD + DB - BC,$ which means that $ BU = DV.$

analytical version on an ellipse with focuses $F_1,F_2$,there are two points $A,B$,the incenters of triangles $AF_jB$ are $I_j$(j=1,2),then $I_1I_2,AB,F_1F_2$ are concurrent.

We first introduce a lemma : $ ABCD $ is a quadrilateral , there is a circle tangent to it whose centre is constructed by intersecting the internal angle bisectors of $ \angle A ~,~ \angle C$ and external angle bisectors of $ \angle B ~,~ \angle D $ if and only if $ AB + BC = CD + DA $ . The proof for only if part is easy , for the if part , construct points $ P,Q $ respectively on $ AB , AD $ away from $A$ such that $ BP = BC ~,~ DQ =DC$ . Then $ AP = AQ $ and therefore the bisectors intersect at the circumcentre of $ \Delta PCQ $ . Now , considering diagonal $BD$ , we just need to show that there is such a circle ( internal bisectors of $ \angle A ~,~ \angle C $ ) iff the concurrence . Let $ I_1 ~,~ I_2 $ be the incentres of $ \Delta ABD ~,~ \Delta BCD $ , $ X = I_1 I_2 \cap AC $ , $ Y = AI_1 \cap CI_2 $ . Be applying Menelaus theorem on $ \Delta I_1 I_2 Y $ and line $ AXC $ , $ \frac{I_2X}{I_1X} = \frac{AY}{AI_1}\cdot \frac{CI_2}{CY} = \frac{r_2}{r_1}\cdot \frac{d(Y,AB)}{d(Y,BC)} $ , where $ r_1 , r_2 $ are the inradii . Therefore , $ d(Y,AB) = d(Y,BC) $ iff $ \frac{I_2X}{I_1X} = \frac{r_2}{r_1} $ , at the same time $ BD $ cuts the segment $ I_1 I_2 $ at the ratio $ r_1 : r_2 $ , we see that what's next is trivial .

here is my solution let the two triangles be $ABD$and $BCD$ .if the two tangency points are symmetical with respect to the midpoint of the diagonal,then we have $AB+BC=AD+DC$,letthe incenter of the triangle $ABD$ and $BCD$ be $I_{1},I_{2}$ respectively, then if $I_{1},I_{2}$ and the intersection of two diagonals are collinear. then we have $\frac{BI_{1}*BI_{2}*sinI_{1}BI_{2}}{DI_{1}*DI_{2}*sinI_{1}DI_{2}}=\frac{AB*BC*sinABC}{AD*DC*sinADC}$ hences to $tan\frac{\angle BDA}{2}tan\frac{\angle BDC}{2}=tan\frac{\angle ABD}{2}tan\frac{\angle DBC}{2}$ so $(AB+BD-AD)(BD+BC-DC)=(BD+DC-BC)(AD+BD-AB)$ so we get $AB+BC=AD+DC$