Very nice problem. Let $O_1$ and $O_2$ denote the circumcenter of $\triangle ATC$ and $BTD$ respectively. Since $ABCD$ is rhombus we clearly have $O_1 \in BD$ and $O_2 \in AC$. Moreover, points $O_1, O_2, T$ are collinear as $\odot(ATC)$ and $\odot(BTD)$ are tangent at point $T$. Observe the following: $$\angle BO_2D = 360^{\circ}-2 \angle BTD = 2\angle ATC =\angle AO_1C$$. This implies that $\angle O_1AC = \angle O_2BD= \angle O_1CA = \angle O_2DB$, hence quadrilaterals $ABO_1O_2$ and $DCO_1O_2$ are cyclic. In particular this gives us $\angle BAO_2 = \angle BO_1T = \alpha$ and $\angle ABO_1 = \angle O_1O_2C = \beta$. Let $X$ and $Y$ be projections of point $T$ on the lines $AC$ and $BD$ respectively and $M$ intersection of $AC$ and $BD$. Observe that: $$ \frac{TY}{TX} = \frac{TO_1 \sin \alpha}{TO_2 \sin \beta} =\frac{AO_1 \sin \alpha}{BO_2 \sin \beta} = \frac{AM \sin \alpha}{MB \sin \beta} =1 $$So $TX=TY$ as needed.

Attachments:

Here's another solution. Claim 1: $\angle ATD=45^\circ$. Proof: Let diagonals $AC,BD$ meet at $O$. Since circles $(ATC)$ and $(BTD)$ are tangent to each other, if $(\ell)$ is their common tangent at $T$ and $K$ a random point on $\ell$, then $\angle CTD=\angle KTD-\angle KTC=\angle TBD-\angle TAC$. Therefore, $\angle CTD=\angle TBD-\angle TAC$ and $\angle ATC=180^\circ-\angle BTD=\angle TBD+\angle TDB$, and so $\angle ATD=\angle ATC-\angle CTD=(\angle TBD+\angle TDB)-(\angle TBD-\angle TAC)=\angle TDB+\angle TAC=\angle AOD-\angle ATD,$ and since $\angle AOD=90^\circ,$ we infer that $\angle ATD=45^\circ,$ as desired $\blacksquare$ Now, let $S,P$ be the circumcenters of triangles $BTC$ and $ATD$, respecively. Claim 2: Triangles $BSC$ and $APD$ are right and isosceles. Moreover, points $S,P$ are symmetric around $O$. Proof: Both triangles are evidently isosceles. Moreover, note that $\angle BTC=\angle BTA+\angle ATC=\angle BTA+180^\circ-\angle BTD=180^\circ-\angle ATD=135^\circ,$ and so $\angle APD=2\angle ATD=90^\circ$ and $\angle BSC=360^\circ-2\angle BTC=90^\circ,$ as desired. For the second part of the Claim, note that triangles $ABC$ and $ACD$ are equal and triangles $BSC$ and $APD$ are equal too, so the Claim follows by symmetry $\blacksquare$ To the problem, note that triangle $STP$ is isosceles, and $O$ is its midpoint, and so $TO \perp SP$, hence $\angle TOB=90^\circ-\angle SOB=90^\circ-\angle BCS=45^\circ,$ and similarly $\angle TOC=45^\circ,$ and so $OT$ bisects angle $\angle BOC$, which is equivalent to saying that $T$ is equidistant from diagonals $AC$ and $BD$.

very nice problem! we denote circumcenter of $\triangle{ATC}$ as $O_{1}$ lies on $BD$ and of $\triangle{BTD}$ as $O_{2}$ lies on $AC$ and intersection of diagonals $AC$ and $BD$ as $M$ and as we have $\odot(ATC)$ and $\odot(BTD)$ to be tangent at point $T$ we have $O_{1},O_{2},T$ to be collinear we Claim:-$ABO_{1}O_{2}$ is cyclic Proof:-we have $\angle{O_{1}AC}=90^{\circ}-\frac{\angle{AO_{1}C}}{2}=90^{\circ}-\angle{ATC}=90^{\circ}-(180^{\circ}-\angle{BTD})=\angle{BTD}-90^{\circ}=\frac{360^{\circ}-\angle{BO_{2}D}}{2}-90^{\circ}=\angle{O_{2}BD}$ hence we have $ABO_{1}O_{2}$ to be concylic $\square$ now we denote $X$ and $Y$ as projections of $T$ onto the lines $AC$ and $BD$ denote $\angle{BO_{1}T}=\alpha , \angle{O_{1}O_{2}C}=\beta$ so and $\angle{O_{2}BM}=\theta$ we have $\frac{BM}{BO_{2}}=\cos{\theta}$ and $\frac{AM}{AO_{1}}=\cos{\theta}$ so we have $\frac{AO_{1}}{BO_{2}}=\frac{AM}{BM}$ and we also have $AO_{1}=TO_{1}$ and $BO_{2}=TO_{2}$ so we have $\frac{TY}{TX}=\frac{TO_{1}\sin{\alpha}}{TO_{2}\sin{\beta}}=\frac{AO_{1}\sin{\alpha}}{BO_{2}\sin{\beta}}=\frac{AM\sin{\alpha}}{BM\sin{\beta}}=1$ hence $TX=TY$ as desired $\blacksquare$

Attachments:

Consider Diagram Let $X = TA \cap (BTD)$ and $Y = TC \cap (BTD)$ Now as both are tangent to each other there exit homothety send $AC$ to $XY$ with ratio $\frac{XY}{AC}$ Let $O$ be center of $(BTD)$ then from anlge condition we get $\angle XOY = \angle BOD \implies XY=BD$ so $\frac{XY}{AC} = \frac{BD}{AC}$ $$[BTD] = \frac{BT.TD.sin\angle BTD}{2}$$$$[ATC] = \frac{AT.TC.sin\angle ATC}{2}$$ $$\frac{[BTD]}{[ATC]} = \frac{BD^2}{AC^2}$$ but area is also base x height, As base are same we get height also in same ratio as ratio in $XY$ and $AC$ which give us $T$ is equal distance from $AC$ and $BD$

Attachments: