In ΔPKO and ΔJKN length of PK= length of KN....given
length of KJ=length of KO.....given∠
measure of ∠PKO=measure of ∠JKN........vertical angles are equalΔ
ΔPKO ≅ ΔJKN SAS
measure of ∠POK=measure of ∠NJO......corresponding parts of congruent triangles
PO is parallel to JN ....alternate-interior angles are equal
An exactly similar argument on the other 2 triangles will get you PJ parallel to ON