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.
afrah , 10 Years ago
Grade 12
1 Answers
SHAIK AASIF AHAMED
Last Activity: 10 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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