Question

2. В треугольнике ABC проведены биссектрисы АК и СL, пересекаю щиеся в точке О. Может ли угол АОС оказаться острым? пожалуйста ​

Answer 1

Первый способ. Пусть ABC - данный треугольник, AM и CK - его биссектрисы (см. рис.). Пусть ∠AOC - острый, то есть, ∠AOC<90∘, тогда, рассмотрев сумму углов треугольника AOC, получим, что ∠OAC+∠OCA>90∘. Следовательно, ∠BAC+∠BCA>180∘, что невозможно, так как это углы треугольника ABC. Второй способ. Пусть ∠ABC=β, тогда вычислим ∠AOC:∠BAC+∠BCA=180∘−β;∠OAC+∠OCA=90∘−0,5β>90∘, то есть, ∠AOC - тупой. Ответ: нет, не может.

Answer 2

Can angle AOS be acute in triangle ABC?

In triangle ABC, let AK and CL be the angle bisectors that intersect at point O. The question is whether angle AOS can be acute.

To determine this, we need to consider the properties of angle bisectors and the relationship between angles in a triangle.

The angle bisector of a triangle divides the opposite side into two segments that are proportional to the adjacent sides. This property is known as the Angle Bisector Theorem.

Let's assume that angle AOS is acute. In this case, angle AOC would also be acute because they are vertical angles.

Since angle AOC is acute, the opposite side AC would be longer than the adjacent sides AO and OC. However, according to the Angle Bisector Theorem, the angle bisector CL divides the side AC into two segments that are proportional to the adjacent sides AO and OC.

Therefore, if angle AOS is acute, it would contradict the Angle Bisector Theorem because the side AC would not be divided proportionally by the angle bisector CL.

Based on this reasoning, we can conclude that angle AOS cannot be acute in triangle ABC.