62nd Czech and Slovak Mathematical Olympiad

Denote the lengths of $AB$, $AD$, and $BD$ by $a$, $b$, and $c$, respectively. If $a=b$ then both $OS$ and $CT$ coincide with $AC$ and the conclusion is trivial. Suppose $a>b$ (the case $b>a$ being completely analogous). Let $T^{\prime }$ be the reflection of $T$ in $S$ (Fig. 1). As $CT\parallel A{T}^{\prime }$, it suffices to prove $OS\parallel A{T}^{\prime }$. Denoting by $E$ the intersection of $AO$ and the diagonal $BD$ we may as well prove $$\frac{AO}{OE}=\frac{T^{\prime }S}{SE}(1)$$ (note that since $a>b$, points $T^{\prime }$, $S$, $E$, and $T$ lie on the diagonal $BD$ in this order). We express both ratios in terms of $a,b,c$. Fig. 1 First, it is well-known that $$DT=\frac{b+c-a}{2},\text{and hence}{T}^{\prime }S=TS=\frac{c}{2}-\frac{b+c-a}{2}=\frac{a-b}{2}.$$ Next, the Angle Bisector Theorem in triangles $ABD$ and $AED$ implies $$BE:ED=AB:AD\text{and}AO:OE=AD:DE$$ which in turn gives $$\begin{gathered} BE=\frac{ac}{a+b}\text{and}DE=\frac{bc}{a+b}, \\ SE=BE-BS=\frac{ac}{a+b}-\frac{c}{2}=\frac{c(a-b)}{2(a+b)}, \\ \frac{AO}{OE}=\frac{AD}{DE}=\frac{b}{\frac{bc}{a+b}}=\frac{a+b}{c}. \end{gathered}$$ Finally for the right-hand side we calculate $$\frac{T^{\prime }S}{SE}=\frac{\frac{a-b}{2}}{\frac{c(a-b)}{2(a+b)}}=\frac{a+b}{c}$$ which finishes the proof.