Solution:

With reference to the accompanying figure, let us draw the line $r$ through $D$ parallel to the side $AB$. Let $L$ and $M$, respectively, be the intersections of $r$ with the sides $AC$ and $BC$. Let us also denote by $H$ and $K$, respectively, the points of tangency of $\gamma$ with the sides $AC$ and $BC$. Since the segments between a given external point and the respective points of contact on the tangents drawn from a point external to a circle are equal (hereafter "the tangent theorem"), we have $CH = CK$, $LH = LD$ and $MK = MD$. Writing $CH = CL + LH$, $CK = CM + MK$ and, using the previous equalities, we get
$$CL + LD = CM + MD.$$

The triangles $CLM$ and $CAB$ are similar, because they have parallel sides. Multiplying the previous equality by the ratio of similarity, we obtain $CA + AS = CB + BS$, that is, $CH + HA + AT + TS = CK + KB + BS$. Using again the tangent theorem, we have $CH = CK$, $AH = AT$ and $KB = TB = TS + SB$. It follows that $2AT + TS = TS + 2BS$, and hence $AT = BS$.