Question 81
Let $ABC$ be a triangle and let $\gamma$ be the circumference inscribed in $ABC$. The circle $\gamma$ is tangent to the side $AB$ at the point $T$. Let $D$ be the point of $\gamma$ diametrically opposite to $T$, and let $S$ be the point of intersection of the straight line passing through $C$ and $D$ with the side $AB$.
Prove that $AT = SB$.
Prove that $AT = SB$.