a)Xét \(\Delta BCE\) và \(\Delta DBE\) có:
\(\widehat{BCE}\)=\(\widehat{BDE}\)(=90\(^0\))
\(\widehat{E}\)chung
=>\(\Delta BCE\)~\(\Delta DBE\)(g.g)
b)Theo câu a)\(\Delta BCE\)~\(\Delta DBE\)=>\(\widehat{CBE}\)=\(\widehat{BDE}\)
Xét \(\Delta BCH\) và \(\Delta BDC\) có:
\(\widehat{CBE}\)=\(\widehat{BDE}\)(cmt)
\(\widehat{CHB}\)=\(\widehat{DCB}\)(=90\(^0\))
=>\(\Delta BCH\)~\(\Delta BDC\)(g.g)
=>\(\dfrac{BC}{BD}=\dfrac{BH}{BC}\)
=>BC\(^2\)=BH.BD