Question

ВАЖНО!!! В треугольнике ABC пусть угол ACB = 120 °, а биссектрисы внутреннего угла, проходящие через A, B и C, пересекают противоположные стороны в A 'или B' или C '. Найти угол A'C'B ? Примечание: правильность результата должна быть доказана.​

Answer 1

∠DCB =180°-120° =60°

∠C1CB =120°/2 =60°

Следовательно CB - биссектриса ∠C1CD

A1 - точка пересечения внешней и внутренней биссектрис △ACC1

В треугольнике биссектрисы двух внешних углов и биссектриса внутреннего угла, не смежного с ними, пересекаются в одной точке, которая является центром вневписанной окружности.

C1A1 - биссектриса ∠CC1B

Аналогично C1B1 - биссектриса ∠CC1A

Биссектрисы смежных углов перпендикулярны, ∠A1C1B1 =90°

Answer 2

Problem Analysis

We are given a triangle ABC with angle ACB measuring 120°. The bisectors of the internal angles passing through points A, B, and C intersect the opposite sides at points A', B', and C'. We need to find the measure of angle A'C'B.

Solution

To find the measure of angle A'C'B, we can use the angle bisector theorem. According to this theorem, the ratio of the lengths of the segments formed by the angle bisector on the opposite sides of the triangle is equal to the ratio of the lengths of the other two sides of the triangle.

Let's denote the length of segment AC' as x and the length of segment BC' as y. Then, we can express the ratios as follows:

AC' / BC' = AC / BC

Since angle ACB measures 120°, we can use the Law of Cosines to express AC and BC in terms of the side lengths of the triangle ABC:

AC = √(AB^2 + BC^2 - 2 * AB * BC * cos(120°)) BC = √(AB^2 + AC^2 - 2 * AB * AC * cos(120°))

Substituting these expressions into the ratio equation, we get:

x / y = √(AB^2 + BC^2 - 2 * AB * BC * cos(120°)) / √(AB^2 + AC^2 - 2 * AB * AC * cos(120°))

Simplifying further, we have:

x / y = √(AB^2 + BC^2 + AB * BC) / √(AB^2 + AC^2 + AB * AC)

Now, let's consider the triangle A'C'B. The sum of the angles in a triangle is always 180°. Therefore, we can express angle A'C'B as:

angle A'C'B = 180° - angle A'CB - angle A'BC

Since the bisectors of the internal angles pass through points A, B, and C, we know that angle A'CB and angle A'BC are both right angles (90°). Therefore, we have:

angle A'C'B = 180° - 90° - 90° = 0°

Hence, the measure of angle A'C'B is 0°.

Conclusion

The measure of angle A'C'B is 0°.

Note: The result has been derived using the angle bisector theorem and the properties of triangles.