a,Xét $ΔBEA$ và $ΔBFH$ có:
$\widehat{BEA}=\widehat{BFH}=90^o$
$\widehat{B}$ chung
$⇒ΔBEA \sim ΔBFH(g.g)⇒\dfrac{BE}{BF}=\dfrac{BA}{BH}$
b,
Xét $ΔBEF$ và $ΔBAH$ có:
$\dfrac{BE}{BF}=\dfrac{BA}{BH}$
$\widehat{C}$ chung
$⇒ΔBEF \sim ΔBAH(c.g.c)$
$⇒\widehat{BEF}=\widehat{BAH}$
c Xét $ΔHBC$ và $ΔABC$ có chung đáy $BC$ đường cao $HD;AD$
$⇒\dfrac{S_{HBC}}{S_{ABC}}=\dfrac{HD}{AD}$
Chứng minh tương tự ta có:
$\dfrac{S_{HAC}}{S_{ABC}}=\dfrac{HE}{BE}$
$\dfrac{S_{HAB}}{S_{ABC}}=\dfrac{HF}{CF}$
$⇒\dfrac{HD}{AD}+\dfrac{HE}{BE}+\dfrac{HF}{CF}=\dfrac{S_{HBC}}{S_{ABC}}+\dfrac{S_{HAC}}{S_{ABC}}+\dfrac{S_{HAB}}{S_{ABC}}=\dfrac{S_{ABC}}{S_{ABC}}=1 (đpcm)$
