In ∆ABC, OB and OC are the bisector of angle ABC and angle ACB respectively. AO produced meets BC at point P. Show that AP is bisector of angle BAC.
Answers 1
Answer:∠BOC = 90 - 1/2
Step-by-step explanation:
∠CBE = 180 - ∠ABC
∠CBO = 1/2 ∠CBE (BO is the bisector of ∠CBE)
∠CBO = 1/2 ( 180 - ∠ABC) 1/2 x 180 = 90
∠CBO = 90 - 1/2 ∠ABC (1) 1/2 x ∠ABC = 1/2∠ABC
∠BCD = 180 - ∠ACD
∠BCO = 1/2 ∠BCD ( CO is the bisector os ∠BCD)
∠BCO = 1/2 (180 - ∠ACD)
∠BCO = 90 - 1/2∠ACD (2)
∠BOC = 180 - (∠CBO + ∠BCO)
∠BOC = 180 - (90 - 1/2∠ABC + 90 - 1/2∠ACD)
∠BOC = 180 - 180 + 1/2∠ABC + 1/2∠ACD
∠BOC = 1/2 (∠ABC + ∠ACD)
∠BOC = 1/2 ( 180 - ∠BAC) (180 -∠BAC = ∠ABC + ∠ACD)
∠BOC = 90 - 1/2∠BAC
Hence proved
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Author:
kahlua
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