Question

Две прямые, проходящие через точку М, лежащую вне окружности с центром О, касаются окружности в точках А и В. Отрезок ОМ делится окружностью пополам. В каком отношении отрезок ОМ делится прямой АВ? Решил, но доказать кое-что не могу

Answer 1

См. рисунок в приложении
АМ=МВ- по свойству касательных проведенных из одной точки
ОА⊥AM
OB⊥BM
Треугольники ОАМ и  ОВМ - прямоугольные
ОА=ОВ=R
ОС=R
По условию
ОС=СM
Значит ОМ=2R
В проямоугольном треугольнике ОАM катет ОА равен половине гипотенузы ОM, значит угол АМО равен 30°.
Угол АОМ равен 60°
Проведем АВ. Хорда АВ в точке К делится пополам ( треугольники АОК и ВОК равны по двум сторонам и углу между ними: АО=ОВ; ОК - общая,

∠АОМ=∠ВОМ = 60°), значит хорда перепендикулярна радиусу ОС

Треугольник АОК - прямоугольный и ∠ОАК=30°
ОК=R/2 
КМ=2R-(R/2)=3R/2
ОК:КМ=R/2 : (3R/2)=1:3

Answer 2

Problem Analysis

We are given that two lines passing through point M, which is outside a circle with center O, are tangent to the circle at points A and B. The segment OM divides the circle into two equal parts. We need to determine the ratio in which segment OM divides the line AB.

Solution

To solve this problem, we can use the property that the tangent to a circle is perpendicular to the radius drawn to the point of tangency. Let's consider the following steps:

1. Draw a diagram to visualize the problem. Place the circle with center O, point M outside the circle, and draw the tangents from point M to the circle, intersecting the circle at points A and B. Also, draw the segment OM and the line AB.

2. Since OM divides the circle into two equal parts, it must pass through the center O. Therefore, OM is a radius of the circle.

3. Let's denote the point of intersection of OM and AB as P. We need to find the ratio in which segment OM divides the line AB, which is equivalent to finding the ratio of lengths MP:PA.

4. Since OM is a radius of the circle, it is perpendicular to the tangent lines MA and MB. Therefore, triangle OMA and triangle OMB are right triangles.

5. Using the property of right triangles, we can apply the Pythagorean theorem to triangles OMA and OMB to relate the lengths of the sides.

6. Let's denote the radius of the circle as r. From the Pythagorean theorem, we have: - In triangle OMA: OA^2 = OM^2 + MA^2 - In triangle OMB: OB^2 = OM^2 + MB^2

7. Since OM divides the circle into two equal parts, we know that OA = OB = r.

8. Substituting OA = OB = r in the above equations, we get: - r^2 = OM^2 + MA^2 - r^2 = OM^2 + MB^2

9. Since we are given that OM divides the circle into two equal parts, we can write: - MA = MB

10. Substituting MA = MB in the above equations, we get: - r^2 = OM^2 + MA^2 - r^2 = OM^2 + MA^2

11. Equating the right-hand sides of the above equations, we have: - OM^2 + MA^2 = OM^2 + MA^2

12. Simplifying the equation, we get: - 0 = 0

13. The equation 0 = 0 is always true, which means that the lengths of MP and PA are equal. Therefore, segment OM divides the line AB in the ratio 1:1.

Answer

Segment OM divides the line AB in the ratio 1:1. This means that the lengths of MP and PA are equal.