a)
Xét $\Delta ADB$ và $\Delta BCD$, ta có:
$\widehat{DAB}=\widehat{CBD}=90{}^\circ $
$\widehat{ABD}=\widehat{BDC}$ ( cùng phụ $\widehat{ADB}$ )
$\to \Delta ADB\backsim\Delta BCD\,\,\,\left( g.g \right)$
b)
Vì $\Delta ADB\backsim\Delta BCD\,\,\,\left( cmt \right)$
$\to \dfrac{AB}{BD}=\dfrac{BD}{DC}$
$\to B{{D}^{2}}=AB.DC$