Lời giải:
Xét tam giác $CDH$ và $CBK$ có:
$\widehat{CHD}=\widehat{CKB}=90^0$
$\widehat{CDH}=180^0-\widehat{ADC}=180^0-\widehat{ABC}=\widehat{CBK}$
$\Rightarrow \triangle CDH\triangle CBK$ (g.g)
$\Rightarrow \frac{CH}{CK}=\frac{CD}{CB}=\frac{AB}{CB}(1)$
và $\widehat{C_1}=\widehat{C_2}$
Có:
$\widehat{KCH}=\widehat{BCD}+\widehat{C_1}+\widehat{C_2}$
$=180^0-\widehat{ABC}+2\widehat{C_2}$
$=180^0-\widehat{ABC}+2(90^0-\widehat{CBK})$
$=360^0-\widehat{ABC}-2\widehat{CBK}$
$=360^0-\widehat{ABC}-2(180^0-\widehat{ABC})$
$=\widehat{ABC}$
Xét tam giác $ABC$ và $HCK$ có:
$\widehat{ABC}=\widehat{HCK}$ (cmt)
$\frac{AB}{BC}=\frac{HC}{CK}$ (theo $(1)$)
$\Rightarrow \triangle ABC\sim \triangle HCK$ (c.g.c)
