Giải thích các bước giải:

a.Xét $\Delta BHK,\Delta ABQ$ có:
Chung $\hat B$
$\widehat{BKH}=\widehat{BQA}(=90^o)$
$\to \Delta BHK\sim\Delta BAQ(g.g)$
b.Từ câu a 
$\to \dfrac{BH}{BA}=\dfrac{BK}{BQ}$
$\to BH\cdot BQ=BK\cdot BA$
c.Xét $\Delta CHQ,\Delta CAK$ có:
Chung $\hat C$
$\widehat{CQH}=\widehat{CKA}(=90^o)$
$\to\Delta CHQ\sim\Delta CAK(g.g)$
$\to \dfrac{CH}{CA}=\dfrac{CQ}{CK}$
$\to CH\cdot CK=CA\cdot CQ$
d.Ta có: $BQ\perp AC, CK\perp AB, BQ\cap CK=H\to H$ là trực tâm $\Delta ABC$
Gọi $AH\cap BC=D\to AD\perp BC$
Xét $\Delta BKC,\Delta BDA$ có:
Chung $\hat B$
$\widehat{BKC}=\widehat{BDA}(=90^o)$
$\to \Delta BKC\sim\Delta BDA(g.g)$
$\to \dfrac{BK}{BD}=\dfrac{BC}{BA}$
$\to BK\cdot BA=BD\cdot BC$
Tương tự $CA\cdot CQ=CD\cdot CB$
$\to BK\cdot BA+CQ\cdot CA=BD\cdot BC+CD\cdot BC=BC(CD+DC)=BC\cdot BC=BC^2$