\(\widehat{BAD}=\widehat{B}+\widehat{C}\)
\(\widehat{ABD}=\frac{180^o-\widehat{BAD}}{2}=90^o-\frac{\widehat{B}+\widehat{C}}{2}\)
\(\widehat{CBD}=\widehat{B}+\widehat{ABD}=\widehat{B}+90^o-\frac{\widehat{B}+\widehat{C}}{2}=90^o+\frac{\widehat{B}-\widehat{C}}{2}=90^o+\frac{\alpha}{2}\)