Giải thích các bước giải:
a.Ta có $BD\perp AC, CE\perp AB, CE\cap BD=H\to H$ là trực tâm $\Delta ABC$
$\to AH\perp BC$
Gọi $AH\cap BC=F$
$\to \widehat{BFH}=\widehat{BDC}=90^o,\widehat{HBF}=\widehat{DBC}$
$\to\Delta BFH\sim\Delta BDC(g.g)$
$\to \dfrac{BF}{BD}=\dfrac{BH}{BC}$
$\to BF.BC=BH.BD$
Tương tự $CH.CE=CF.CB$
$\to BH.BD+CH.CE=BF.BC+CF.CB=BC^2$
b.Ta có: $\widehat{AEC}=\widehat{ADB}=90^o,\widehat{EAC}=\widehat{BAD}$
$\to \Delta ADB\sim\Delta AEC(g.g)$
$\to \dfrac{AD}{AE}=\dfrac{AB}{AC}$
$\to\dfrac{AD}{AB}=\dfrac{AE}{AC}$
Mà $\widehat{EAD}=\widehat{BAC}$
$\to \Delta ADE\sim\Delta ABC(c.g.c)$
$\to\widehat{ACB}=\widehat{AED}=40^o$
$\to\widehat{DCB}=40^o$
$\to \widehat{DBC}=90^o-\widehat{DCB}=50^o$
$\to\widehat{HBC}=50^o$
c.Ta có: $\widehat{AKB}=90^o$
$\to \widehat{AEK}=\widehat{AKB}=90^o$
Mà $\widehat{EAK}=\widehat{BAK}$
$\to \Delta AEK\sim\Delta AKB(g.g)$
$\to \dfrac{AE}{AK}=\dfrac{AK}{AB}$
$\to AK^2=AE.AB$
Tương tự $AI^2=AD.AC$
Từ câu b ta có: $\dfrac{AE}{AC}=\dfrac{AD}{AB}\to AE.AB=AD.AC$
$\to AK^2=AI^2$
$\to AK=AI$
