1. **Problem Statement:** Determine if triangle $\triangle NMB$ is semi-equilateral given the angles and side lengths in $\triangle ANB$ and the geometric relationships described. 2. **Given Information:** - $\triangle ANB$ with $\angle ANB = 90^\circ$ (angle facing diameter). - $\angle ABN = 30^\circ$, $\angle NAB = 60^\circ$. - Side $AB = 8$ cm. - $AN$ is opposite $30^\circ$, $NB$ opposite $60^\circ$. 3. **Calculate sides in $\triangle ANB$:** - Hypotenuse $AB = 8$ cm. - Side opposite $30^\circ$ (AN) = $\frac{hyp}{2} = \frac{8}{2} = 4$ cm. - Side opposite $60^\circ$ (NB) = $\frac{hyp \sqrt{3}}{2} = \frac{8 \sqrt{3}}{2} = 4\sqrt{3}$ cm. 4. **Check if $\triangle ANB$ is semi-equilateral:** - A semi-equilateral triangle has angles $30^\circ$, $60^\circ$, and $90^\circ$ with sides in ratio $1 : \sqrt{3} : 2$. - $AN : NB : AB = 4 : 4\sqrt{3} : 8$ simplifies to $1 : \sqrt{3} : 2$. - So, $\triangle ANB$ is semi-equilateral. 5. **Analyze $\triangle NMB$:** - Point $M$ is above $B$ forming a right angle at $B$ with $BM \perp AB$. - Since $\angle ANB = 90^\circ$ and $\angle OSB = 90^\circ$, and $AN \parallel OM$ by converse of corresponding angles, $\triangle NMB$ shares similar angle properties. 6. **Conclusion:** - $\triangle NMB$ has a right angle at $B$ and shares the angle $30^\circ$ at $N$ (since $\angle ABN = 30^\circ$ and $M$ lies on the perpendicular at $B$). - Therefore, $\triangle NMB$ also has angles $30^\circ$, $60^\circ$, and $90^\circ$. - Hence, $\triangle NMB$ is semi-equilateral. **Final answer:** Yes, $\triangle NMB$ is semi-equilateral.