6. As shown in Figure $3, A B$ is the diameter of the semicircle $\odot O$, $C$ is a moving point on the semicircle arc (excluding the midpoint of the semicircle arc and the two endpoints of the diameter), connect $A C, B C$, draw $C D \perp A B$ at point $D$, draw $\odot O_{1}$ tangent to $A D$ at point $E$, tangent to $C D$ and $\odot O$, draw $\odot O_{2}$ tangent to $D B$ at point $F$, tangent to $C D$ and $\odot O$. Then the degree of $\angle E C F$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
6. B.

Connect $O O_{1}$ and extend it, then this line segment must pass through the tangent point G. Thus,
$$
A O=O B=O G=R \text {. }
$$

Let $A D=m, D B=n, m+n=2 R$.
Connect $O_{1} E$, then $O_{1} E \perp A D$.
In the right triangle $\triangle O_{1} E O$,
$$
O_{1} O^{2}=O_{1} E^{2}+O E^{2} \text {. }
$$

Also, $O_{1} O=R-r_{1}, O_{1} E=r_{1}$,
$$
O E=E D-(A D-A O)=r_{1}-(m-R) \text {. }
$$

Substitute into equation (1) and simplify to get
$$
r_{1}^{2}+2 n r_{1}+\left(n^{2}-2 R n\right)=0 \text {. }
$$

Solving for $r_{1}$ gives $r_{1}=\sqrt{2 R n}-n$.
By the problem, $C D^{2}=A D \cdot D B=m n$,
$$
\begin{array}{l}
A C=\sqrt{C D^{2}+A D^{2}}=\sqrt{m n+m^{2}} \\
=\sqrt{m(n+m)}=\sqrt{2 R m} .
\end{array}
$$

Also, $A E=m-r_{1}=2 R-\sqrt{2 R n}$, then
$$
\frac{A C}{C D}=\frac{\sqrt{2 R m}}{\sqrt{m n}}=\sqrt{\frac{2 R}{n}}=\frac{2 R-\sqrt{2 R n}}{\sqrt{2 R n}-n}=\frac{A E}{E D} \text {. }
$$

Therefore, $C E$ bisects $\angle A C D$.
Similarly, $C F$ bisects $\angle B C D$.
Thus, $\angle E C F=\frac{1}{2}(\angle A C D+\angle B C D)=45^{\circ}$.