The perpendicular dropped from the top of the parallelogram to the diagonal divides it into segments 6 and 15

The perpendicular dropped from the top of the parallelogram to the diagonal divides it into segments 6 and 15. Find the sides and diagonals of the parallelogram if it is known that the difference of the sides is 7.

In the parallelogram AB and BC – sides, BC = AB + 7; VN – perpendicular to the AC diagonal;

AH = 6; HC = 15;

From triangle ABN: BH² = AB² – AH²;

From triangle BHC: BH² = BC² – HC²;

AB² – AH² = BC² – HC²;

AB² – AH² = (AB + 7) ² – HC²; AB² – AH² = AB² + 14 AB + 49 – HC²;

14 AB = HC² – AH² – 49 = 225 – 36 – 49 = 140;

AB = 140: 14 = 10; BC = 10 + 7 = 17;

Diagonal AC = AH + HC = 6 + 15 = 21;

BC diagonal:

BC² = 2 AB² + 2 BC² – AC² = 2 * 10² + 2 * 17² – 21² = 200 + 289 – 441 = 49;

BC = √49 = 7.