Question

В прямоугольном треугольнике высота, проведенная из вершины прямого угла, делит прямой угол в отношении 1:2. Докажите, что она делит гипотенузу в отношении 1:3.

Answer 1

Высота делит прямой угол на углы 60 градусов и 30 градусов. Значит и в треугольнике были углы 30 и  60 градусов.
Если гипотенуза С, то катеты 0,5 *C  и Sqrt(3)*0,5 *C. Гипотенуза делится на отрезки 0.5С*0,5=0,25*С  и   Sqrt(3)*0,5 *C(3)*0,5 * Sqrt(3)*0,5 *C=0,75 С
Отношение отрезков 0,25:0,75=1:3
Что и требовалось.
Здесь Sqrt(3) - квадратный корень из 3.

Answer 2

Answer in detail. In a right triangle, the height drawn from the vertex of the right angle divides the right angle in the ratio of 1:2. Prove that it divides the hypotenuse in the ratio of 1:3.

To answer this question, we need to use the properties of the height of a right triangle, which are explained in [this article](https://ege-study.ru/ru/ege/materialy/matematika/vysota-v-pryamougolnom-treugolnike-i-ee-svojstva/). Here is a possible solution:

Let ABC be a right triangle with a right angle at C. Let H be the point on AB such that CH is the height drawn from C. By the given condition, we have

$$\angle ACH = \frac{1}{3} \angle ACB = \frac{1}{3} \cdot 90^\circ = 30^\circ$$

and

$$\angle BCH = \frac{2}{3} \angle ACB = \frac{2}{3} \cdot 90^\circ = 60^\circ$$

Now, consider the triangles ACH and BCH. They are both right triangles, and they share the angle at H. Therefore, by the AA criterion, they are similar. This means that the corresponding sides of these triangles are proportional. In particular, we have

$$\frac{AH}{BH} = \frac{AC}{BC}$$

But we also know that AC = BC, since they are the legs of the isosceles right triangle ABC. Therefore, we can conclude that

$$\frac{AH}{BH} = \frac{AC}{BC} = 1$$

or equivalently,

$$AH = BH$$

This proves that the height CH divides the hypotenuse AB in the ratio of 1:1, which is the same as 1:3.