Giải thích các bước giải:
a.Ta có $AB\perp AC, HE\perp AB,HF\perp AC\to AEHF$ là hình chữ nhật
b.Ta có:$\widehat{AEH}=\widehat{AHB}=90^o,\widehat{EAH}=\widehat{BAH}$
$\to \Delta AEH\sim\Delta AHB(g.g)$
$\to \dfrac{AE}{AH}=\dfrac{AH}{AB}$
$\to AH^2=AE.AB$
Tương tự $AH^2=AF.AC$
$\to AE.AB=AF.AC=AH^2$
c.Ta có $\widehat{BEH}=\widehat{BHA}=90^o,\widehat{EBH}=\widehat{ABH}$
$\to \Delta BEH\sim\Delta BHA(g.g)$
$\to \dfrac{BE}{BH}=\dfrac{BH}{BA}$
$\to BE=\dfrac{BH^2}{BA}$
Tương tự $CF=\dfrac{CH^2}{CA}$
$\to \dfrac{EB}{FC}=\dfrac{BH^2}{BA}:\dfrac{CH^2}{CA}$
$\to \dfrac{EB}{FC}=\dfrac{BH^2}{CH^2}\cdot \dfrac{CA}{BA}$
Ta có $\widehat{AHB}=\widehat{BAC}=90^o,\widehat{ABH}=\widehat{ABC}$
$\to \Delta BHA\sim\Delta BAC(g.g)$
$\to \dfrac{BH}{BA}=\dfrac{BA}{BC}$
$\to BH=\dfrac{BA^2}{BC}$
Tương tự $CH=\dfrac{CA^2}{CB}$
$\to \dfrac{BH^2}{CH^2}=\dfrac{AB^4}{AC^4}$
$\to \dfrac{EB}{FC}=\dfrac{AB^3}{AC^3}$
d.Từ câu c ta có:
$EB=\dfrac{BH^2}{BA},CF=\dfrac{CH^2}{CA}$
$\to BE\cdot CF=\dfrac{BH^2\cdot CH^2}{BA\cdot CA}$
$\to BE\cdot CF=\dfrac{(BH\cdot CH)^2}{HA\cdot BC}$ vì $AH\cdot BC=AB\cdot AC=2S_{ABC})$
Ta có $\widehat{AHB}=\widehat{AHC}=90^o,\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to \Delta AHB\sim\Delta CHA(g.g)$
$\to \dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to AH^2=HB\cdot HC$
$\to BE\cdot CF=\dfrac{AH^4}{AH\cdot BC}$
$\to BC.BE.CF=AH^3$
