РЕШЕНИЕ:



AD=8, BC=6
MD = 8 – 6 = 2

∆CDM ∞ ∆FCB
MD : BC = CD : FC
2 : 6 = x : FC
FC = 3x

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

∆FKE (∠K = 90^o)
EK = FE ∙ cos E = FE ∙ cos D = = FE ∙ MD/CD = 2 x ∙ 2/x = 4

Ответ: 4

РЕШЕНИЕ:



AD=10, BC=8
MD = 10 – 8 = 2

∆CDM ∞ ∆FCB
MD : BC = CD : FC
2 : 8 = x : FC
FC = 4x

FE² = FD ∙ FC
FE² = (FC + CD) ∙ FC
FE² = 5x ∙ x
FE = √5 x

∆FKE (∠K = 90^o)
EK = FE ∙ cos E = FE ∙ cos D = = FE ∙ MD/CD = √5 x ∙ 2/x = 2√5

Ответ: 2√5

РЕШЕНИЕ:



AD=15, BC=12
MD = 15 – 12 = 3

∆CDM ∞ ∆FCB
MD : BC = CD : FC
3 : 12 = x : FC
FC = 4x

FE² = FD ∙ FC
FE² = (FC + CD) ∙ FC
FE² = 5x ∙ x
FE = √5 x

∆FKE (∠K = 90^o)
EK = FE ∙ cos E = FE ∙ cos D = = FE ∙ MD/CD = √5 x ∙ 3/x = 3√5

Ответ: 3√5

РЕШЕНИЕ:



AD=24, BC=12
MD = 24 – 12 = 12

∆CDM ∞ ∆FCB
MD : BC = CD : FC
12 : 12 = 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 ∙ 12/x = 12√2

Ответ: 12√2

РЕШЕНИЕ:



AD=6, BC=3
MD = 6 – 3 = 3

∆CDM ∞ ∆FCB
MD : BC = CD : FC
3 : 3 = 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 ∙ 3/x = 3√2

Ответ: 3√2

РЕШЕНИЕ:



AD=15, BC=14
MD = 15 – 14 = 1

∆CDM ∞ ∆FCB
MD : BC = CD : FC
1 : 14 = x : FC
FC = 14x

FE² = FD ∙ FC
FE² = (FC + CD) ∙ FC
FE² = 15x ∙ x
FE = √15 x

∆FKE (∠K = 90^o)
EK = FE ∙ cos E = FE ∙ cos D = = FE ∙ MD/CD = √15 x ∙ 1/x = √15

Ответ: √15