20 DH-?

В прямоугольном треугольнике ADH, $$DH^2 = AD^2 - AH^2$$. $$AH = AB - HB = 8 - x$$. $$DH = 8 an(60^ ext{o}) = 8 ext{sqrt}(3)$$. $$AD^2 = AH^2 + DH^2$$. $$AD^2 = (8-x)^2 + (8 ext{sqrt}(3))^2$$. $$BC = AH + HB = 8-x+x = 8$$. $$CD = AB = 8$$. $$AD = BC = 8$$. $$8^2 = (8-x)^2 + (8 ext{sqrt}(3))^2$$. $$64 = 64 - 16x + x^2 + 192$$. $$x^2 - 16x + 192 = 0$$. $$DH = 8 an(60^ ext{o}) = 8 ext{sqrt}(3)$$.
Ответ: $$8 ext{sqrt}(3)$$