Given a right-angled triangle ABC with legs BC = 3 and AC = 4. Rhombus BDEF is located in triangle

Given a right-angled triangle ABC with legs BC = 3 and AC = 4. Rhombus BDEF is located in triangle ABC, vertex B is common, and the other three vertices of the rhombus lie on three sides of triangle ABC. Find the side of the rhombus.

By the Pythagorean theorem, we find the hypotenuse AB as the root of the sum of the squares of the legs: AB = √ (AC ^ 2 + BC ^ 2) = √ (4 ^ 2 + 3 ^ 2) = √ (16 + 9) = √25 = 5.
The ratio of the adjacent leg to the hypotenuse is equal to the cosine of the angle: cos B = BC / AB = 3/5 = 0.6.
If the vertex B is common, then the two sides of the rhombus adjacent to it belong to the sides of the triangle. Let the side of the rhombus BF be part of the leg BC, and the side of the rhombus BD be part of the hypotenuse AB, the apex of the rhombus E lies on the side of the triangle AC. The opposite sides of the diamond ED and BF are parallel, therefore the side of the diamond ED is parallel to the side of the triangle CB. Triangles AED and ACB are similar because vertex A is common, sides AE and AC, AD and AB coincide with each other, so the angles B and ADE are equal, then cos B = cos ADE = 0.6.
Let the side of the rhombus be x, then from the triangle AED:
ED / AD = cos ADE;
ED / (AB-BD) = x / (AB-x) = x / (5-x) = 0.6;
x = (5-x) * 0.6;
x = 3-0.6x;
1.6x = 3;
x = 3 / 1.6 = 1.875.
The desired side of the rhombus is 1.875.