We know by sine rule ; $\frac{V_c}{Sin[90-(\Phi-\alpha)]}=\frac{V_f}{Sin\Phi}=\frac{V_s}{Sin(90-\alpha)}$

Also $r_c=\frac{t_2}{t_1}=\frac{cos(\Phi-\alpha)}{sin\Phi}$

using; $\frac{V_c}{Sin[90-(\Phi-\alpha)]}=\frac{V_f}{Sin_\Phi}$

Therefore,$V_c=\frac{V_f}{Sin\Phi}\times Sin[90-(\Phi-\alpha)]$

but,$sin[90-(\Phi-\alpha)]=cos(\Phi-\alpha)$

Therefore $V_c=V_f \times r_c$

Hence proved