Question

В треугольнике АВС медиана АD и бессектриса ВЕ пересекаются в точке О. Если AD перпердикулярно ВЕ и S(AOE)=2,то площадь треугольника АВС равнаесли можно с рисунком

Answer 1

Рисунок не могу. А вы нарисуйте по моему решению :)
Пока НЕ пользуюсь перпендикулярностью биссектрисы и медианы.
Хотя наоборот гораздо легче. 
Пусть х = ВD/АВ; 
AE/AC = AE/(AE + EC) = 1/(1 + EC/AE) = 1/(1 + BC/AB) = 1/(1 + 2*BD/AB) = 1/(1 + 2*x);
Тогда Sabe = Sabc*AE/AC = Sabc/(1 + 2*x);AO/AD = AO/(AO + OD) = 1/(1 + OD/AO) = 1/(1 + BD/AB) = 1/(1 + x);
Saob = Sadb*AO/AD = Sadb/(1 + x);Sadb = Sabc/2; (AD -медиана)
=> Saob = Sabc/(2 + 2*x) Поэтому
2 = Saoe = Saeb - Saob = Sabc*(1/(1 + 2*x) - 1/(2 + 2*x));
(Вот только теперь стукнем себя по лбу и воскликнем:))) Раз в треугольнике АВD биссектриса перпендикулярна основанию, то это равнобедренный треугольник. Поэтому AB = BD, х = 1;
2 = Sabc*(1/3 - 1/4) = Sabc/12;  Sabc = 24;

Задача делалась давно, но раз я ошибся, есть долг :) Вот другое решение, основанное на том, что с самого начала очевидно, что биссектриса BO - высота в треугольнике ABD, то есть AB = BD = BC/2; 
На продолжении BA за точку A я отмечаю точку F, так, что AF = AB;Очевидно, что AD II FC; AD - средняя линяя в треугольнике FBC; FD, AC и BE - медианы в треугольнике FBC; 
Отсюда следует вот что
1) Площадь треугольника FBC Sfbc = 2*Sabc; (AC - медиана! :) )
2) Медианы делят треугольник на 6 равных по площади треугольников, то естьSabe = Sfbc/6 = Sabc/3;
3) Средняя линия отсекает от треугольника подобный ему треугольник с вдвое меньшими сторонами, то есть Sabd = Sfbc/4; => Saob = Sabd/2 = Sfbc/8 = Sabc/4; откуда 
2 = Sabc*(1/3 - 1/4) = Sabc/12; Sabc = 24; 

Answer 2

Triangle ABC with Median AD and Bisector BE

In triangle ABC, the median AD and the bisector BE intersect at point O. It is given that AD is perpendicular to BE and the area of triangle AOE is 2. We need to find the area of triangle ABC.

To solve this problem, we can use the properties of medians and bisectors in a triangle.

Properties of Medians in a Triangle

A median of a triangle is a line segment that connects a vertex to the midpoint of the opposite side. Here are some important properties of medians:

1. The medians of a triangle intersect at a point called the centroid. In this case, point O is the centroid of triangle ABC.

2. The centroid divides each median into two segments. The segment from the centroid to the vertex is twice as long as the segment from the centroid to the midpoint of the opposite side.

Properties of Bisectors in a Triangle

A bisector of a triangle is a line segment that divides an angle into two equal angles. Here are some important properties of bisectors:

1. The bisectors of a triangle intersect at a point called the incenter. In this case, point O is the incenter of triangle ABC.

2. The incenter is equidistant from the sides of the triangle. This means that the distances from point O to the sides AB, BC, and AC are equal.

Solution

Since AD is perpendicular to BE, we can conclude that triangle AOE is a right triangle. The area of a right triangle can be calculated using the formula:

Area = (1/2) * base * height

In triangle AOE, the base is AE and the height is AD. Given that the area of triangle AOE is 2, we can write:

2 = (1/2) * AE * AD

Since AD is a median, we know that AE = 2 * EO. Therefore, we can rewrite the equation as:

2 = (1/2) * (2 * EO) * AD

Simplifying the equation, we get:

2 = EO * AD

Now, let's consider triangle ABC. The area of triangle ABC can be calculated using the formula:

Area = (1/2) * base * height

In triangle ABC, the base is BC and the height is AD. We need to find the length of BC.

Since AD is a median, we know that BD = DC. Therefore, BC = 2 * BD.

Let's substitute the values into the formula:

Area = (1/2) * BC * AD

Substituting BC = 2 * BD, we get:

Area = (1/2) * (2 * BD) * AD

Since BD = DC, we can rewrite the equation as:

Area = (1/2) * (2 * DC) * AD

Simplifying the equation, we get:

Area = DC * AD

From the given information, we know that the area of triangle AOE is 2, which means EO * AD = 2. Substituting this value into the equation, we get:

Area = DC * 2

Therefore, the area of triangle ABC is 2 times the length of DC.

Unfortunately, without specific measurements or additional information about the triangle, we cannot determine the exact value of the area of triangle ABC. However, we can conclude that the area of triangle ABC is directly proportional to the length of DC.

I apologize for not being able to provide a more precise answer without further information or a diagram. If you have any additional details or measurements, please provide them, and I will be happy to assist you further.