Question

In triangle ABC the sides AB and AC are produced to point E and D respectively if bisector BO and CO of angle CBE and angle BCD respectively meet at point o then prove that *angle BOC=90-1/2 angle A* .

Answer 1

Sol:


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