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1. 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 Pand meets MK produced at Q. Prove that (i) PKMN is a cyclic quadrilateral, (ii) NP=NM.

1. 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 Pand meets MK produced at Q.
Prove that
(i) PKMN is a cyclic quadrilateral,

(ii) NP=NM.

Grade:12

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
Please find the answer to your question below
ang. LKM =90° and ang PNM =90°
we know that
opposite angles of a cyclic quadrilateral are supplementary
so PKMN is a cyclic quadrilateral.
Similarly we can show that NP=NM but here as there is no diagram provided i could not visualize it.

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