$\triangle ABC$ có $AB< AC$, ba đường cao $AE, BD, CF$ cắt nhau tại $H$
Lời giải:
a) Xét $\triangle BEH$ và $\triangle BDC$ có:
$\begin{cases}\widehat{B}:\ \text{góc chung}\\\widehat{E} = \widehat{D} = 90^\circ\end{cases}$
Do đó: $\triangle BEH\backsim \triangle BDC\ (g.g)$
$\Rightarrow \dfrac{BE}{BD} = \dfrac{BH}{BC}$
$\Rightarrow BH.BD = BE.BC$
Xét $\triangle CEH$ và $\triangle CFB$ có:
$\begin{cases}\widehat{C}:\ \text{góc chung}\\\widehat{E} = \widehat{F} = 90^\circ\end{cases}$
Do đó: $\triangle CEH\backsim \triangle CFB\ (g.g)$
$\Rightarrow \dfrac{CE}{CF} = \dfrac{CH}{CB}$
$\Rightarrow CH.CF = CE.BC$
Khi đó ta được:
$BH.BD + CH.CF$
$= BE.BC + CE.BC$
$= BC(BE + CE)$
$= BC.BC$
$= BC^2$
b) Ta có:
$\dfrac{HE}{AE} =\dfrac{\dfrac12HE\cdot BC}{\dfrac12AE\cdot BC} = \dfrac{S_{BHC}}{S_{ABC}}$
Tương tự:
$\dfrac{HD}{BD} = \dfrac{S_{AHC}}{S_{ABC}}$
$\dfrac{HF}{CF} = \dfrac{S_{AHB}}{S_{ABC}}$
Cộng vế theo vế ta được:
$\quad \dfrac{HE}{AE} + \dfrac{HD}{BD} + \dfrac{HF}{CF} = \dfrac{S_{BHC}}{S_{ABC}}+\dfrac{S_{AHC}}{S_{ABC}}+\dfrac{S_{AHB}}{S_{ABC}}$
$\Leftrightarrow \dfrac{HE}{AE} + \dfrac{HD}{BD} + \dfrac{HF}{CF} =\dfrac{S_{BHC} + S_{AHC} + S_{AHB}}{S_{ABC}}$
$\Leftrightarrow \dfrac{HE}{AE} + \dfrac{HD}{BD} + \dfrac{HF}{CF} = \dfrac{S_{ABC}}{S_{ABC}}$
$\Leftrightarrow \dfrac{HE}{AE} + \dfrac{HD}{BD} + \dfrac{HF}{CF} = 1$
