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: $\dfrac{AE}{AC}=\dfrac{AD}{AB}\to AE.AB=AD.AC$

$\to AK^2=AI^2$

$\to AK=AI$