Question

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

Answer 1

Answer:

Tyehjdvnsnjx hwuhegwhiudggs bsyyehksnhrhkevhdjjh jwuyegg3uudhejhsge ejjehuehdhb3 heihrhhemk7tuu7

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).

Learn more :-

PQR is an isosceles triangle in which PQ=PR. Side QP is produced to such that PS=PQ Show

that QRS is a right angle

https://brainly.in/question/23326569

In triangle ABC, if AL is perpendicular to BC and AM is the bisector of angle A. Show that angle LAM= 1/2 ( angle B - an...

https://brainly.in/question/2117081