Đáp án+Giải thích các bước giải:

a)Xét $\triangle$DHB và $\triangle$EHC có:

$\widehat{DBH}$=$\widehat{CEH}$(=$90^o$)

$\widehat{BHD}$=$\widehat{CHE}$(2 góc đối đỉnh)

$\Rightarrow$$\triangle$DHB$\backsim$$\triangle$EHC(g-g)

b)Xét $\triangle$ABE và $\triangle$ACD có:

$\widehat{AEB}$=$\widehat{ADC}$(=$90^o$)

$\widehat{BAC}$ chung

$\Rightarrow$$\triangle$ABE$\backsim$$\triangle$ACD(g-g)

$\Rightarrow$ $\dfrac{AE}{AD}$=$\dfrac{AB}{AC}$

$\Rightarrow$ AE.AC=AB.AD

c)Vì BE$\bot$AC

CD$\bot$AB

mà BD$\cap$CD=H

$\Rightarrow$ H là trực tâm

$\Rightarrow$ AH$\bot$BC hay AF$\bot$BC

Ta có :$\dfrac{ sBHC}{sABC}$=$\dfrac{\dfrac{1}{2}HF.BC}{\dfrac{1}{2}AF.BC}$=$\dfrac{HF}{AF}$(1)

$\dfrac{ sAHB}{sABC}$=$\dfrac{\dfrac{1}{2}HD.AB}{\dfrac{1}{2}CD.AB}$=$\dfrac{HD}{CD}$(2)

$\dfrac{ sAHC}{sABC}$=$\dfrac{\dfrac{1}{2}HE.AC}{\dfrac{1}{2}BE.AC}$=$\dfrac{HE}{BE}$(3)

Từ (1)(2)(3)

$\Rightarrow$$\dfrac{HF}{AF}$ + $\dfrac{HD}{CD}$ +$\dfrac{HE}{BE}$

=$\dfrac{ sBHC}{sABC}$+$\dfrac{ sAHB}{sABC}$+$\dfrac{ sAHC}{sABC}$

=$\dfrac{ sBHC+sAHB+sAHC}{sABC}$=$\dfrac{ sABC}{sABC}$=1(Đpcm)