c) Chứng minh: BH.BD + CH.CF = BC2 và HE/AE + HD/BD+ HF/CF=1

c) +) Xét hai tam giác DBEH và DBDC có:

$\left.\begin{array}{c}\widehat{EBH}=\widehat{DBC}\left(\widehat{B}chung\right)\\ \widehat{BEH}=\widehat{BDC}\left(=90°\right)\end{array}\right\}\Rightarrow \Delta BEH\backsim \Delta BDC\left(g.g\right)$

$\Rightarrow \frac{BE}{BD}=\frac{BH}{BC}\iff BH.BD=BE.BC$ (1)

+) Xét hai tam giác DCEH và DCFB có:

$\left.\begin{array}{c}\widehat{ECH}=\widehat{FCB}\left(\widehat{C}chung\right)\\ \widehat{CEH}=\widehat{CFB}\left(=90°\right)\end{array}\right\}\Rightarrow \Delta CEH\backsim \Delta CFB\left(g.g\right)$

$\Rightarrow \frac{CE}{CF}=\frac{CH}{CB}\iff CH.CF=CE.CB$ (2)

Từ (1) và (2) ta có:

BH.BD + CH.CF = BE.BC + CE.BC

= BC(BE + CE) = BC.BC = BC² (đpcm)

+) Ta có:

$\frac{HE}{AE}+\frac{HD}{BD}+\frac{HF}{CF}$

 $=\frac{\frac{1}{2}.HE.BC}{\frac{1}{2}AE.BC}+\frac{\frac{1}{2}.HD.AC}{\frac{1}{2}.BD.AC}+\frac{\frac{1}{2}.HF.AB}{\frac{1}{2}.CF.AB}$

$=\frac{S_{HBC}}{S_{ABC}}+\frac{S_{HAC}}{S_{BAC}}+\frac{S_{HAB}}{S_{CAB}}$

$=\frac{S_{HBC}+S_{HAC}+S_{HAB}}{S_{ABC}}=\frac{S_{ABC}}{S_{ABC}}=1$ (đpcm).