Giải:
1. Ta có:
$\widehat{AHE} = \widehat{AHN} + \widehat{EHN} = \widehat{ABN} + \widehat{EMN} = \widehat{ABN} + \widehat{EBN} = 90^{\circ}$
Vậy tứ giác AHEK nội tiếp.

2. Ta có:
$\widehat{NFK} = \widehat{NFA} + \widehat{AFK} = \widehat{NMA} + \widehat{ABK} = \widehat{NMA} + \widehat{EBK} = \widehat{NMA} + \widehat{EMK} = \widehat{NME} = \widehat{NEM}$
Vậy tam giác NFK cân.

Ta có:
$\frac{EM}{EN} = \frac{EM}{EC} \cdot \frac{EC}{EN} = \frac{sin\widehat{EMC}}{sin\widehat{ENC}} \cdot \frac{sin\widehat{NEC}}{sin\widehat{MEC}} = \frac{sin\widehat{EMC}}{sin\widehat{MEC}} \cdot \frac{sin\widehat{NEC}}{sin\widehat{ENC}} = \frac{sin\widehat{EMC}}{sin\widehat{MEC}} \cdot \frac{sin\widehat{NEC}}{sin\widehat{ENC}} = \frac{sin\widehat{EMC}}{sin\widehat{MEC}} \cdot \frac{sin\widehat{NEC}}{sin\widehat{ENC}} = \frac{CM}{CN}$
Vậy $EM.NC = EN.CM$

3. Giả sử $KE = KC$. Ta có:
$\widehat{EKN} = \widehat{EKC} + \widehat{CKN} = \widehat{EKC} + \widehat{CBN} = \widehat{EKC} + \widehat{EBN} = \widehat{EKC} + \widehat{EMN} = \widehat{EKM}$
Vậy $OK // MN$