a)
Xét $\Delta AHB$ và $\Delta DAB$, ta có:
$\widehat{AHB}=\widehat{DAB}=90{}^\circ $
$\widehat{ABH}$ là góc chung
$\Rightarrow \Delta AHB\backsim\Delta DAB\,\,\,\left( g.g \right)$
b)
Xét $\Delta HBI$ và $\Delta HKD$, ta có:
$\widehat{BHI}=\widehat{KHD}=90{}^\circ $
$\widehat{HBI}=\widehat{HKD}$ ( cùng phụ $\widehat{BDC}$ )
$\Rightarrow \Delta HBI\backsim\Delta HKD\,\,\,\left( g.g \right)$
$\Rightarrow \dfrac{HB}{HK}=\dfrac{HI}{HD}$
$\Rightarrow HB.HD=HI.HK$
c)
$\Delta AHD$ có $BI\,//\,AD\,\,\,\Rightarrow \dfrac{AH}{AI}=\dfrac{DH}{DB}$ ( định lý Ta-let )
$\Delta ABH$ có $DK\,//\,AB\,\,\,\Rightarrow \dfrac{AH}{AK}=\dfrac{BH}{BD}$ ( định lý Ta-let )
Cộng vế theo vế, ta được:
$\,\,\,\,\,\,\,\dfrac{AH}{AI}+\dfrac{AH}{AK}=\dfrac{DH}{DB}+\dfrac{BH}{BD}$
$\Rightarrow AH\left( \dfrac{1}{AI}+\dfrac{1}{AK} \right)=\dfrac{DH+BH}{BD}$
$\Rightarrow AH\left( \dfrac{1}{AI}+\dfrac{1}{AK} \right)=\dfrac{BD}{BD}$
$\Rightarrow AH\left( \dfrac{1}{AI}+\dfrac{1}{AK} \right)=1$
$\Rightarrow \dfrac{1}{AI}+\dfrac{1}{AK}=\dfrac{1}{AH}$
