Для начала найдем векторы:

AM = AB + BM = b/2 + b/2 = b

DN = DM + MN = DC/2 + AN = c/2 + b/2

Теперь разложим вектор DN по векторам b,d и c:

DN = (DN • b) b/ |b| ^ 2 + (DN • d) d/ |d| ^ 2 + (DN • c) * c/ |c| ^ 2

где (DN • b), (DN • d) и (DN • c) - скалярные произведения векторов DN и b,d,c соответственно.

(DN • b) = (c/2 + b/2) • b = (c/2) • b + (b/2) • b = (c/2) |b| cos(180) + (b/2) |b| cos(0) = -c/2 |b| + b/2 |b| = -c/2 b + b/2 b = (b^2-c^2)/2

(DN • d) = (c/2 + b/2) • d = (c/2) • d + (b/2) • d = 0

(DN • c) = (c/2 + b/2) • c = (c/2) • c + (b/2) • c = (c/2) |c| cos(0) + (b/2) |c| cos(90) = (c/2) * |c| + 0 = c^2/2

Итак, разложение вектора DN по векторам b,d и c:

DN = (b^2-c^2)/2 b/ |b| ^ 2 + 0 d/ |d| ^ 2 + c^2/2 c/ |c| ^ 2 = (b^2-c^2)/(2|b|^2) b + (c^2)/(2|c|^2) * c

Ответ: DN = (b^2-c^2)/(2|b|^2) b + (c^2)/(2|c|^2) c