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.
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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.
we know that
opposite angles of a cyclic quadrilateral are supplementary
so PKMN is a cyclic quadrilateral.
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