Question

Triangle KLM in which the side KL has greater length than KM, ∠LKM = 90° and the bisector of ∠LKM meets LM at N. The line through N perpendicular to LM cuts KL at P and meets MK produced at Q.
Prove that
   (i) PKMN is a cyclic quadrilateral,
  (ii) NP = NM.

Answer 1

1.    ang. LKM =  90°  and ang PNM =  90°
  we know that
   opposite angles of a cyclic quadrilateral are supplementary
  so PKMN is a cyclic quadrilateral.