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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 equals 90° plus one by two angle a?
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Understanding the Triangle and Angle Bisectors
In triangle ABC, let the angles be defined as follows:
- Angle A = a
- Angle B = b
- Angle C = c
The angle bisectors of angles B and C meet at point O.
Using Angle Relationships
To demonstrate that angle BOC = 90° + 1/2 angle A, we can use the properties of angles in a triangle:
- The sum of angles in triangle ABC is:
a + b + c = 180°
- Therefore, we can express angle A in terms of the other angles:
a = 180° - (b + c)
Calculating Angle BOC
To find angle BOC, consider the angles around point O:
- Angle BOC can be expressed as:
BOC = 180° - (1/2 b + 1/2 c)
- This simplifies to:
BOC = 180° - 1/2 (b + c)
- Since a = 180° - (b + c), we substitute:
BOC = 180° - 1/2 (180° - a)
BOC = 180° - 90° + 1/2 a
BOC = 90° + 1/2 a
Conclusion
Thus, we have proven that:
Angle BOC = 90° + 1/2 angle A.
This relationship highlights the special properties of angle bisectors in a triangle, specifically in triangle ABC, confirming that the angles created by the intersection of bisectors retain a predictable relationship with the triangle's angles.
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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 equals 90° plus one by two angle a?
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