in triangle ABC the bisectors of angle B and angle C meet at O.Prove that angle BOC=90 degree+ angle A/2
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Here BO, CO are the angle bisectors of ∠PBC & ∠QCB intersect each othe at O.
∴ ∠1 = ∠2 and ∠3 = ∠4
Side AB and AC of ΔABC are produced to P and Q respectively.
∴ Exterior of ∠PBC = ∠A + ∠C --------------(1)
And Exterior of ∠QCB = ∠A + ∠B --------------(2)
Addiing (1) and (2) we get
∠PBC + ∠QCB = 2 ∠A + ∠B + ∠C.
2∠2 + 2∠3 = ∠A + 180°
∠2 + ∠3 = (1 /2)∠A + 90° ----------(3)
But in a ΔBOC = ∠2 + ∠3 + ∠BOC = 180° --------( 4)
From equ (3) and (4) we get
(1 /2)∠A + 90° + ∠BOC = 180°
∠BOC = 90° - (1 /2)∠A
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Ur answer is here ↪️↪️↪️↪️↪️↪️↪️↪️↪️↪️
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I will be using A, B, C not angle A or angle B.... Hope you will not mind it.
As we know A + B + C = 180
So, B + C = 180 - A
Or, (B/2) + (C/2) = 90 - (A/2) ....(1)
Now think in triangle BOC,
we have three angles, OBC, BOC and OCB
OBC = (B/2)
OCB = (C/2)
Let BOC = X
So, X + (B/2) + (C/2) = 180
From equation (1)
X + 90 - (A/2) = 180
X = 90 + A/2 (Proved !!!)
✴️✴️✴️✴️✴️✴️✴️✴️✴️✴️(-:(-:(-:(-:(-:(-:(-:(-:(-:(-:(-:(-:(-:
. HOPE ITS HELP UH!!!
Ur answer is here ↪️↪️↪️↪️↪️↪️↪️↪️↪️↪️
____________________
I will be using A, B, C not angle A or angle B.... Hope you will not mind it.
As we know A + B + C = 180
So, B + C = 180 - A
Or, (B/2) + (C/2) = 90 - (A/2) ....(1)
Now think in triangle BOC,
we have three angles, OBC, BOC and OCB
OBC = (B/2)
OCB = (C/2)
Let BOC = X
So, X + (B/2) + (C/2) = 180
From equation (1)
X + 90 - (A/2) = 180
X = 90 + A/2 (Proved !!!)
✴️✴️✴️✴️✴️✴️✴️✴️✴️✴️(-:(-:(-:(-:(-:(-:(-:(-:(-:(-:(-:(-:(-:
. HOPE ITS HELP UH!!!
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