Electrical engineering help?

Refer to the figure below.

V(AC) = V(BD) = V(BE) = 20V

The equivalent resistance of R₄ and R₅, R₄₅
= 1 / [(1/8) + (1/8)] kΩ
= 4 kΩ

The equivalent resistance of R₃, R₄ and R₅, R₃₄₅
= (4 + 6) kΩ
= 10 kΩ

The equivalent resistance of the whole circuit, Rₛ
= 1 / [(1/25) + (1/10) + (1/10)] kΩ
= 4.167 kΩ

V(AC) = Iₛ Rₛ
20 V = Iₛ (4.167 kΩ)
Iₛ = 4.800 mA

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b)
V(BE) = I₁ R₃₄₅
20 V = I₁ (10 kΩ)
I₁ = 2 mA

V₁ = I₁ R₄₅
V₁ = (2mA) × (4 kΩ)
V₁ = 8.000 V

OR:
V₁ = Vₛ × (R₄₅/R₃₄₅)
V₁ = (20 V) × (4/10)
V₁ = 8.000 V

8//8=4+6=10//10=5//25=4166.67ohms
I=20/4166.67=.0048A
v1=(20*4000)/10000=8.000v