Giải thích các bước giải:
a.Ta có $\widehat{CHI}=\widehat{CAB}=90^o,\widehat{ICH}=\widehat{ACB}$
$\to\Delta HIC\sim\Delta ABC(g.g)$
b. Ta có $\widehat{IAK}=\widehat{IHC}=90^o,\widehat{AIK}=\widehat{CIH}$
$\to\Delta IAK\sim\Delta IHC(g.g)$
$\to\dfrac{IA}{IH}=\dfrac{IK}{IC}$
$\to IA.IC=IH.IK$
c.Ta có $IH\perp BC, BA\perp CI, IH\cap BA=K\to K$ là trực tâm $\Delta IBC$
$\to CK\perp BI$
$\to\widehat{BDC}=\widehat{BHI}=90^o,\widehat{DBC}=\widehat{IBH}$
$\to\Delta BDC\sim\Delta BHI(g.g)$
$\to\dfrac{BD}{BH}=\dfrac{BC}{BI}$
$\to BD.BI=BH.BC$
Tương tự chứng minh được $CA.CI=CH.CB$
$\to BD.BI+CA.CI=BH.BC+CH.CB=BC^2$
d.Ta có $\widehat{BDK}=\widehat{KAC}=90^o,\widehat{DKB}=\widehat{AKC}$
$\to\Delta DKB\sim\Delta AKC(g.g)$
$\to\dfrac{KD}{KA}=\dfrac{KB}{KC}$
$\to\dfrac{KD}{KB}=\dfrac{KA}{KC}$
Mà $\widehat{DKA}=\widehat{BKC}$
$\to\Delta KDA\sim\Delta KBC(c.g.c)$
$\to\widehat{KDA}=\widehat{KBC}=\widehat{ABC}$
