BMO 2017

We will prove first that the circle $c_1$ is tangent to $AB$ at the point $B$. In order to prove this, we have to prove that $\angle BDC=\angle ABC$. Indeed, since $BD\parallel AC$, we have that $\angle DBC=\angle ACB$. Additionally, $\angle BCD=\angle BAC$ (by chord and tangent), which means that the triangles $ABC,BDC$ have two equal angles and so the third ones are also equal. It follows that $\angle BDC=\angle ABC$, so $c_1$ is tangent to $AB$ at the point $B$.

Similarly, the circle $c_2$ is tangent to $AC$ at the point $C$.

As a consequence, $\angle ABT=\angle ACB$ (by chord and tangent) and also $\angle BSC=\angle ACB$.

By the above, we have that $\angle ABT=\angle BSC$, so the lines $BT,SC$ are parallel.

Now, let $ST$ intersect $BC$ at the point $K$. It suffices to prove that $K$ belongs to $AL$.

From the trapezoid $BTCS$ we get that $$\frac{BK}{KC}=\frac{BT}{SC}(1)$$ and from the similar triangles $ABT,ASC$, we have that $$\frac{BT}{SC}=\frac{AB}{AS}(2).$$ By (1), (2) we get that $$\frac{BK}{KC}=\frac{AB}{AS}(3).$$ From the power of point theorem, we have that $$A{C}^2=AB\cdot AS\Rightarrow AS=\frac{A{C}^2}{AB}.$$ Going back into (3), it gives that $$\frac{BK}{KC}=\frac{A{B}^2}{A{C}^2}.$$ From the last one, it follows that $K$ belongs to the symmedian of the triangle $ABC$.

Finally, recall that the well known fact that since $LB$ and $LC$ are tangents, it follows that $AL$ is the symmedian of the triangle $ABC$, so $K$ belongs to $AL$, as needed.