Question

O is any point inside a triangle pqr the bisectors of angle poq, angle qor and angle por meet the sides pq, qr and pr in points a, B, and C respectively. Prove that ap bq cr =aq br Cp

Answer 1

AQ * BR * CP = AP * BQ * CR  pqr the bisectors of angle poq, angle qor and angle por meet the sides pq, qr and pr in points a, B, and C respectively.

Step-by-step explanation:

Lets draw a Line parallel to QO passing through R such that it intersect extended OB at M

=> RM ║ QO

=> ∠QOB = ∠RMB

  ∠QOB = ∠ROB   ( OB is bisector of ∠POR)

=> ∠RMB = ∠ROB

=> RO = RM

in ΔPOB  & Δ MRB

∠QOB = ∠RMB

∠QBO = ∠RBM ( oppsite angles)

∠OQB = ∠MRB

=> ΔPOB  ≈ Δ MRB

=> QO/RM  = QB/BR

=> QO/RO = QB/BR

Similarly

   OP/OQ = PA/AQ

   OR/OP = RC/CP

Multiplying all three

(OQ/RO)(OP/OQ)(OR/OP)  = (QB/BR)(PA/AQ)(RC/CP)

=> AQ * BR * CP = AP * BQ * CR

QED

Proved

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Answer 2

answer = ✓ hope it will help you !