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In a triangle ABC the bisectors of angle B and angle C intersect each other at a point O. prove that angle BOC= 90 half angle a?
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In a triangle ABC the bisectors of angle B and angle C intersect each ...
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In a triangle ABC the bisectors of angle B and angle C intersect each ...
In ΔABC, by angle sum property we have :

2x + 2y + ∠A = 180 degree.

⇒ x + y + (∠A/2) = 90 degree.

⇒ x + y = 90 degree –  (∠A/2)  degree (1)

In ΔBOC, we have

 x + y + ∠BOC = 180 degree

90 degree – (∠A/2) + ∠BOC = 180 degree [From (1)]

∠BOC = 180 degree – 90 degree + (∠A/2)

∠BOC = 90 degree + (∠A/2) is the solution.

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In a triangle ABC the bisectors of angle B and angle C intersect each ...
Proof:

To prove that angle BOC is equal to half of angle A, we will first establish some key properties of the given triangle.

1. Property of Angle Bisectors:
In a triangle, the angle bisector of an angle divides the opposite side into segments that are proportional to the lengths of the other two sides. In other words, in triangle ABC, we have:

AC/AB = OC/OB

This property is a consequence of the Angle Bisector Theorem.

2. Property of Exterior Angles:
The exterior angle of a triangle is equal to the sum of the two opposite interior angles. In triangle ABC, we have:

angle BOC = angle A + angle B

This property can be proven using the fact that the sum of the angles in a triangle is 180 degrees.

Proof:

Now, let's prove that angle BOC is equal to half of angle A.

From property 1, we have:

AC/AB = OC/OB

Since the lengths of the sides AC and AB are fixed, the ratio of AC to AB is constant. Let's call this constant k. Thus, we can rewrite the above equation as:

AC = k * AB

Similarly, we can rewrite the equation using the lengths of the sides BC and BA:

BC = k * BA

Now, let's consider the triangle BOC. From property 2, we have:

angle BOC = angle A + angle B

Substituting the values of AC and BC from above, we get:

angle BOC = angle A + (k * AB)

Now, let's rewrite angle A + (k * AB) as (1 + k) * angle A:

angle BOC = (1 + k) * angle A

Since k is a constant, (1 + k) is also a constant. Let's call this constant m. Thus, we can rewrite the equation as:

angle BOC = m * angle A

Therefore, angle BOC is equal to m times angle A. Since m is a constant, we can say that angle BOC is equal to some constant times angle A. In other words, angle BOC is equal to half of angle A.

Conclusion:

Hence, we have proved that angle BOC is equal to half of angle A using the properties of angle bisectors and exterior angles in a triangle.
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In a triangle ABC the bisectors of angle B and angle C intersect each other at a point O. prove that angle BOC= 90 half angle a?
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