1. **Problem statement:** We have a circle with center $T$, $FR$ as diameter, with $FI \perp FM$, $\angle RFM = 30^\circ$, and $\angle IFB = 40^\circ$. We want to find $\angle BFR$. 2. Since $FR$ is a diameter, $\angle FRB$ is a right angle ($90^\circ$) because it subtends a semicircle. 3. Given $FI \perp FM$, so $\angle IFM = 90^\circ$. 4. Note that $\angle RFM = 30^\circ$, and $FI$ is perpendicular to $FM$ at $F$. So, $\angle IFM = 90^\circ$. 5. Consider triangle $FIB$. Since $\angle IFB = 40^\circ$ and $FI$ is perpendicular to $FM$, the remaining angle at $F$ in triangle $BFR$ is what we need to find. 6. Use angle relationships at $F$: - $\angle BFR = \angle RFM + \angle IFB = 30^\circ + 40^\circ = 70^\circ$ 7. Thus, $\boxed{70^\circ}$ is the measure of $\angle BFR$.