Bisectors of angles B and C of a triangle ABC intersect . . .

Question : Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC=90°+∠A/2

OR

In a ΔABC, if the bisectors of ∠B and ∠C meet at point O, then ∠BOC is equal to 90°+∠A/2

Solution : 

In ΔABC

∠A+∠B+∠C=180°(Angle Sum Property)
Multiply both sides by ½
½×∠A+½×∠B+½×∠C=½×180°
∠A/2+∠B/2+∠C/2 =90°
∠A/2+∠B/2+∠C/2 =90°
∠A/2+∠1+∠2 =90°
∠1+∠2 =90°-∠A/2— (1) 

In ΔOBC

∠BOC+∠1+∠2=180° (ASP)
∠BOC+90°-∠A/2=180° [From eq. (1)]
∠BOC=180°-90°+∠A/2

∠BOC=90°+∠A/2

Similar Questions 

1. If one angles of a triangle 130°, then the angle between the bisectors of the other two angles can be 
(a) 50°
(b) 65°
(c) 145°
(d) 155°

2. Assertion (A) : In a ΔABC, if the bisectors of angles of ∠B and ∠C meet at a point O, then ∠BOC is always an obtuse angle. 

Reason (R) : In a ΔABC, if the bisectors of ∠B and ∠C meet at a point O, then ∠BOC=90°+∠A/2

(A) Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.

(B) Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.

(C) Assertion (A) is true, but Reason (R) is false.

(D) Assertion (A) is false, but Reason (R) is true.