Giải thích các bước giải:
Xét $\Delta BEI,\Delta MEC$ có:
$\widehat{IBE}=\widehat{ECM}(=45^o)$
$EB=EC$
$\widehat{IEB}=90^o-\widehat{BEM}=\widehat{MEC}$
$\to\Delta BEI=\Delta CEM(g.c.g)$
$\to BI=CM, EI=EM$
Ta có:
$\widehat{AMB}=\widehat{CMN},\widehat{ABM}=\widehat{MCN}(=90^o)$
$\to\Delta BAM\sim\Delta CNM(g.g)$
$\to \dfrac{BA}{CN}=\dfrac{BM}{CM}$
$\to\dfrac{BC}{CN}=\dfrac{BM}{BI}$
$\to \dfrac{BI}{CN}=\dfrac{BM}{BC}$
Mà $\widehat{IBM}=\widehat{BCN}$
$\to\Delta BIM\sim\Delta CNB(c.g.c)$
Mặt khác $\widehat{IBM}=\widehat{IEM}=90^o\to BIEM$ nội tiếp
$\to \widehat{BNC}=\widehat{BIM}=\widehat{BEM}=\widehat{BEK}$
$\to DNKE$ nội tiếp
$\to \widehat{BKE}=\widehat{BDN}=45^o=\widehat{ECB}$
$\to BECK$ nội tiếp
$\to \widehat{BKC}=180^o-\widehat{BEC}=90^o$
$\to CK\perp BN$
