Xét $\Delta OHA$ và $\Delta OAC$ có:
$\widehat{AHO}=\widehat{CAO}=90^o$ $(AH\perp OC)$
$\widehat{AOH}=\widehat{COA}$ (cùng là 1 góc)
$\to \Delta OHA\sim\Delta OAC$ (g.g)
$\to\dfrac{OA}{OC}=\dfrac{OH}{OA}\to OA^2=OH.OC$
$\to OB^2=OH.OC(AO=OB)$
$\to\dfrac{OB}{OH}=\dfrac{OC}{OB}$
$\to\Delta OBH\sim\Delta OCB(c.g.c)$
$\to \widehat{OHB}=\widehat{OBC}$ (hai góc tương ứng) $=\widehat{ABC}=\widehat{AHD}$ (giả thiết)
$\to \widehat{DHB}=\widehat{BHO}+\widehat{OHD}=\widehat{AHD}+\widehat{OHD}=\widehat{OHA}=90^o$
$\to HB\perp HD$ (đpcm)