Question

please answer this
in the figure
ab and ac are produced to e and d
if bisectors of bo and co of cbe and bcd meet at o. prove that boc = 90-1/2 bac

Answer 1

$\sf{\huge{Heya \: mate.\: Solution\\ below}}$
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$\tt{\red{\bold{\underline{Answer -}}}}$





♠ Firstly, BO is the bisector of ∠CBE.

=> ∠CBO = ∠OBE = ½∠CBE .... (i)



♠ Similarly CO is the bisector of ∠BCD

=> ∠BCO = ∠OCD = ½∠BCD ....(ii)





♠ Now, since ABE is a straight line, so by linear pair,

∠ABC + ∠CBE = 180°


• Multiplying by 1/2 on both sides,

=> (½ × ∠ABC) + (½ × ∠CBE) = (½ × 180°)

=> ½∠B + ½∠CBE = 90°

=> ½∠B + ∠CBO = 90° ...(using eq i)

=> ∠CBO = (90° - ½∠B) ....(iii)




♠ Similarly, since ACD is also a straight line, we can find that,

=> ∠BCO = (90° - ½∠C) ....(iv)




♠ Now, we know that, sum of all angles of a triangle is 180°.

So, in ∆OBC,

∠CBO + ∠BCO + ∠BOC = 180°

=> (90° - ½∠B) + (90° - ½∠C) + ∠BOC = 180°

=> 90° + 90° - ½∠B - ½∠C + ∠BOC = 180°

=> 180° - ½(∠B + ∠C) + ∠BOC = 180°

=> ∠BOC - ½(∠B + ∠C) = 0 ...(cancelling 180°)

=> ∠BOC = ½(∠B + ∠C) ...(shifting to R.H.S)


Now, adding and subtracting ½∠A in L.H.S

=> ∠BOC = ½(∠A + ∠B + ∠C) - ½∠A

=> ∠BOC = ½(180°) - ½∠A ...(sum of all sides = 180°)

=> ∠BOC = 90° - ½∠A

=> ∠BOC = 90° - ½∠BAC



$\huge{\mathcal{Hence \: proved}}$
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Thank you... (^_-)

Answer 2

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