3. Given a circle with center $O$ and diameter $A B$. Point $C$ is chosen on the circle such that $\overline{D B}=3 \overline{O M}$, where $D$ is the projection of $C$ onto the diameter $A B$, and $M$ is the projection of $O$ onto $B C$. Determine $\measuredangle A B C$.
Solution. Let $\overline{O M}=x$ and $\overline{O B}=r$. According to the condition $\overline{D B}=3 \overline{O M}=3 x$, then $\overline{A D}=\overline{A B} - \overline{D B}=2 r-3 x$. From the fact that $O M$ is the midline in $\triangle A B C$, we have $\overline{A C}=2 \overline{O M}=2 x$. The triangles OBM and ÅCD are similar, so $\frac{\overline{A C}}{\overline{A D}}=\frac{\overline{O B}}{\overline{O M}} \quad$ i.e. $\frac{2 x}{2 r-3 x}=\frac{r}{x}$, from which $2 x^{2}+3 r x-2 r^{2}=0$. The solutions to this quadratic

![](https://cdn.mathpix.com/cropped/2024_06_05_d0c212a3407f50323a50g-04.jpg?height=339&width=390&top_left_y=1253&top_left_x=1074)
equation are: $x_{1 / 2}= \pm \frac{r}{2}$, from which $r=2 x$.

Thus, $\overline{A C}=\overline{A O}=\overline{O C}$, so angle $C A B$ is $60^{\circ}$. It follows that the desired angle is $30^{\circ}$.