1. **Problem Statement:** Prove using vectors that the angle subtended by a diameter of a circle at any point on the semicircle is a right angle (90 degrees). 2. **Setup and Notation:** Let the diameter be the segment from $-\mathbf{u}$ to $\mathbf{u}$, so the center is at the origin $\mathbf{0}$. Point $\mathbf{v}$ lies on the semicircle above the diameter. 3. **Key Formula:** The angle $\theta$ at $\mathbf{v}$ between vectors $\mathbf{v} - (-\mathbf{u}) = \mathbf{v} + \mathbf{u}$ and $\mathbf{v} - \mathbf{u}$ satisfies $$\cos \theta = \frac{(\mathbf{v} + \mathbf{u}) \cdot (\mathbf{v} - \mathbf{u})}{|\mathbf{v} + \mathbf{u}| |\mathbf{v} - \mathbf{u}|}$$ 4. **Important Rule:** If $\cos \theta = 0$, then $\theta = 90^\circ$. 5. **Calculate the dot product:** $$ (\mathbf{v} + \mathbf{u}) \cdot (\mathbf{v} - \mathbf{u}) = \mathbf{v} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} - \mathbf{u} \cdot \mathbf{u} $$ Since dot product is commutative, $\mathbf{v} \cdot \mathbf{u} = \mathbf{u} \cdot \mathbf{v}$, so the middle terms cancel: $$ - \mathbf{v} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} = 0 $$ Thus, $$ (\mathbf{v} + \mathbf{u}) \cdot (\mathbf{v} - \mathbf{u}) = |\mathbf{v}|^2 - |\mathbf{u}|^2 $$ 6. **Use the fact that $\mathbf{v}$ lies on the semicircle:** Since the semicircle has radius $|\mathbf{u}|$, we have $$ |\mathbf{v}| = |\mathbf{u}| $$ Therefore, $$ |\mathbf{v}|^2 - |\mathbf{u}|^2 = 0 $$ 7. **Conclusion:** $$ \cos \theta = \frac{0}{|\mathbf{v} + \mathbf{u}| |\mathbf{v} - \mathbf{u}|} = 0 $$ Hence, $$ \theta = 90^\circ $$ which proves the angle in a semicircle is a right angle. **Final answer:** The angle subtended by the diameter at any point on the semicircle is $90^\circ$.