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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.



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.

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