13. In the triangle $A B C, \angle A=20^{\circ}, \angle C=90^{\circ}, O$ is a point on $A B$ and $D$ is the midpoint of $O B$. Suppose the circle centered at $O$ with radius $O D$ touches the side $A C$ at $T$. Determine the size of $\angle B C D$ in degrees.
13. Ans: 35

Join $D T$ and $D C$. Let $M$ be the foot of the perpendicular from $D$ onto $A C$. Then $O T$, $D M$ and $B C$ are all parallel. Since $D$ is the midpoint of $O B, M$ is the midpoint of $T C$. Thus $D T C$ is an isosceles triangle with $D T=D C$ and $\angle T D M=\angle M D C$.

As $\angle T O A=70^{\circ}$ and $\triangle O T D$ is isosceles, we have $\angle O T D=35^{\circ}$. Thus $\angle B C D=$ $\angle M D C=\angle T D M=\angle O T D=35^{\circ}$.