Vì $\widehat{DAB} = \widehat{CBA}$ => $\widehat{HAD} = \widehat{KBC}$ (góc bù nhau)
Trong $\Delta HAD$ và $\Delta KBC$ có:
$\widehat{AHD} = \widehat{BKC} = 90^{0}$
$\widehat{HAD} = \widehat{KBC}$
=> $\widehat{HDA} = \widehat{KCB}$
mà DH = CK
=> $\Delta HAD$ = $\Delta KBC$ (g.c.g)
=> AD = BC