Giải thích các bước giải:
a.Xét $\Delta AHB,\Delta BDE$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BED}(=90^o)$
$\to \Delta BHA\sim\Delta BED(g.g)$
b.Xét $\Delta CHA, \Delta CDF$ có:
chung $\hat C$
$\widehat{CHA}=\widehat{CFD}(=90^o)$
$\to \Delta CHA\sim\Delta CFD(g.g)$
$\to \dfrac{CA}{CD}=\dfrac{HA}{FD}$
$\to AC.DF=DC.AH$
c.Từ câu a
$\to \dfrac{AB}{BD}=\dfrac{HA}{ED}$
$\to AB.DE=AH.BD$
$\to AB.DE+AC.DF=AH.BD+DC.AH=AH(BD+DC)=AH.BC$