Math, asked by pallupradhan162, 1 year ago

In the figure , sides AB and AC of triangle ABC are produced to point E and D respectively . if bisectors BO and CO of angle CBE and angle BCD respectively meet at point O, then prove that angle BOC =90°-1/2 angle BAC . 15 th question

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Answered by Cyra
625

Ray BO is the bisector of angle CBE
Therefore,angle CBO=1/2of angle CBE
=1/2(180-y)
=90-y/2 (1)
Similarly,ray OC is the bisector of angle BCD
Therefore,angle BCO=1/2 of angle BCD
=1/2(180-z)
=90-z/2 (2)
In triangle BOC,angle BOC+BCO+CBO=180 (3)
Substituting (1,2,3) you get
Angle BOC+90-z/2+90-y/2=180
Angle BOC=z/2+y/2
=1/2(y+z)
But,x+y+z=180 (angle sum property)
y+z=180-x
Angle BOC=1/2(180-x)
=90-x/2
=90-1/2angle BAC

Answered by littyissacpe8b60
384

∠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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