a) Vì BH \(\perp\) CD (gt), BD \(\perp\) BC (gt)
=> \(\widehat{H_1}=\widehat{H_2}=\widehat{BDC}=90^o\) (ĐN 2 đường thẳng \(\perp\))
Xét \(\Delta\)HDB và \(\Delta\)BCD có:
\(\widehat{H_1}=\widehat{BCD}\) (cmt)
\(\widehat{D_1}\): chung
=> \(\Delta\)HDB ~ \(\Delta\)BCD (g.g)