РЕШЕНИЕ:



AD=4, BC=3
MD = 4 – 3 = 1

∆CDM ∞ ∆FCB
MD : BC = CD : FC
1 : 3 = x : FC
FC = 3x

FE² = FD ∙ FC
FE² = (FC + CD) ∙ FC
FE² = 4x ∙ x
FE = 2x

∆FKE (∠K = 90^o)
EK = FE ∙ cos E = FE ∙ cos D = = FE ∙ MD/CD = 2x ∙ 1/x = 2

Ответ: 2

РЕШЕНИЕ:



AD=16, BC=8
MD = 16 – 8 = 8

∆CDM ∞ ∆FCB
MD : BC = CD : FC
8 : 8 = x : FC
FC = x

FE² = FD ∙ FC
FE² = (FC + CD) ∙ FC
FE² = 2x ∙ x
FE = √2 x

∆FKE (∠K = 90^o)
EK = FE ∙ cos E = FE ∙ cos D = = FE ∙ MD/CD = √2 x ∙ 8/x = 8√2

Ответ: 8√2