The height of the rhombus divides its side into segments m and n. Find the diagonal of the rhombus.

Let us prove that the triangles AOD and BDН are similar.

Both triangles are rectangular, AOD, since the diagonals at the intersection point form a right angle, BDH, since BH is height. The angle D is common for triangles, then the triangles are similar in acute angle.

Let AC = d1, BD = d ^ 2

Then: BD / AD = DH / OD.

d1 / (m + n) = n / (d1 / 2).

d1 ^ 2/2 = n * (m + n).

d1 = √ (2 * n * (m + n)).

From the right-angled triangle AOD, by the Pythagorean theorem, we define the leg AO.

AO ^ 2 = AD ^ 2 – OD ^ 2.

(d2 / 2) ^ 2 = (m + n) ^ 2 – (d1 / 2) ^ 2.

d2 ^ 2 = 4 * ((m + n) ^ 2 – (√ (2 * n * (m + n)) / 2) ^ 2.

d2 = √ (4 * (m + n) 2) – (2 * n * (m + n))).

Answer: AC = √ (2 * n * (m + n)), BD = √ (4 * (m + n) 2) – (2 * n * (m + n))).