Equate the lengths of the cross products, so that $∥\mathbf{A}∥ ∥\mathbf{B}∥$ $\sin \left({\mathrm{\θ}}_1\right)$ = $∥\mathbf{A}∥$ $∥\mathbf{C}∥$ $\sin \left({\mathrm{\θ}}_2\right)$, where ${\mathrm{\θ}}_1$ is the angle between A and B, and ${\mathrm{\θ}}_2$ is the angle between A and C. This gives

$\frac{∥\mathbf{B}∥}{\∥\mathbf{C}\∥}$ = $\frac{\sin \left({\mathrm{\θ}}_2\right)}{\sin \left({\mathrm{\θ}}_1\right)}$ 

The equality $\mathbf{A}·\mathbf{B}\=\mathbf{A}·\mathbf{C}$ gives $∥\mathbf{A}∥∥\mathbf{B}∥\cos \({\mathrm{\θ}}_1\)$ = $∥\mathbf{A}∥∥\mathbf{C}∥\cos \({\mathrm{\θ}}_2\)$, from which it follows that

$\frac{∥\mathbf{B}∥}{\∥\mathbf{C}\∥}$ = $\frac{\cos \left({\mathrm{\θ}}_2\right)}{\cos \left({\mathrm{\θ}}_1\right)}$

Therefore $\frac{\sin \left({\mathrm{\theta }}_2\right)}{\sin \left({\mathrm{\theta }}_1\right)}$ = $\frac{\cos \left({\mathrm{\theta }}_2\right)}{\cos \left({\mathrm{\theta }}_1\right)}$ so $\sin \left({\mathrm{\θ}}_2\right)\cos \left({\mathrm{\θ}}_1\right)-\cos \left({\mathrm{\θ}}_2\right)\sin \left({\mathrm{\θ}}_1\right)\=0\=\sin \left({\mathrm{\θ}}_2-{\mathrm{\θ}}_1\right)$, from which it follows that ${\mathrm{\θ}}_2\={\mathrm{\θ}}_1$. Consequently, $∥\mathbf{B}∥$ = $∥\mathbf{C}∥$, and since the angles these vectors make with the fixed vector A are the same, the vectors B and C must be the same.$$