Question

Show that in a triangle ABC the exterior angle bisectors of angle B and C inclined at an angle 90-A/2.​

Answer 1

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In △ABC,

∠ABC+∠ACB+∠BAC=180

∠ABC+70+50=180

∠ABC=60∘

∠OCB=1/2(180−∠ACB)

∠OCB=1/2(180−50)

∠OCB=65°

∠OBC=1/2(180−∠ABC)

∠OBC=1/2(180−60)

∠OBC=60

In △OBC,

∠OCB+∠OBC+∠BOC=180

65+60+∠BOC=180

∠BOC=180−125

∴∠BOC=55°

$\huge\mathfrak\red{itz\:jyotsana☺}$

Answer 2

Given :- show that in a triangle ABC the exterior angle bisector of Angle B and C inclined at O, then angle BOC = 90°- (A/2) ?

Solution :-

from image we have :-

• Exterior angle bisector of ∠ABE and Exterior angle bisector of ∠ACF meets at O.

Now,

→ ∠ABE = ∠A + ∠C (Exterior angle is equal to sum of opposite interior angles in a ∆.)

than,

→ ∠EBH = ∠ABH = (1/2)[∠A + ∠C] (angle bisector.)

So,

→ ∠CBO = ∠EBH = (1/2)[∠A + ∠C] (vertically opposite angles.)

Similarly,

→ ∠BCO = = (1/2)[∠A + ∠B]

Now, In Triangle OBC we have ,

→ ∠CBO = (1/2)[∠A + ∠C]

→ ∠BCO = (1/2)[∠A + ∠B]

So,

→ ∠BOC + ∠CBO + ∠BCO = 180° (Angle sum Property.)

→ ∠BOC = 180° - (∠CBO + ∠BCO)

→ ∠BOC = 180° - (1/2)[∠A + ∠C+ ∠A + ∠B]

→ ∠BOC = 180° - (1/2)[∠A + 180°]

→ ∠BOC = 180° - 90° - (1/2)∠A

→ ∠BOC = 90° - (∠A/2) (Proved).

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