Giải thích các bước giải:
a.Ta có $\Delta AHD\sim\Delta BAD$ vì:
Chung $\hat D,\widehat{AHD}=\widehat{DAB}$
$\Delta BAD\sim\Delta DCB$ vì:
$\widehat{ABD}=\widehat{BDC},\widehat{ADB}=\widehat{DBC}$
$\to \Delta AHD\sim\Delta DCB$
$\Delta DCB\sim\Delta BHA$ vì:
$\widehat{BDC}=\widehat{ABH},\widehat{DCB}=\widehat{AHB}(=90^o)$
$\Delta AHB\sim\Delta DAB$ vì:
Chung $\hat B, \widehat{AHB}=\widehat{DAB}$
$\to \Delta AHD\sim\Delta BHA$
b.Từ câu a
$\to \dfrac{AH}{BH}=\dfrac{HD}{HA}\to AH^2=HB.HD$
$\dfrac{AB}{BH}=\dfrac{BD}{AB}\to AB^2=BH.BD$
c.Ta có $BD=\sqrt{AB^2+AD^2}=\sqrt{194}$
Mà $AH.BC=AB.AD=(2S_{ABD})$
$\to AH=\dfrac{AB.AD}{BC}=\dfrac{65}{\sqrt{194}}$