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Let $r$ be the radius of the third circle. Then,

From the above diagram, we get

$AE= GC + CD$

$\Rightarrow 4 = r + CD $

$ \Rightarrow CD = ( 4-r )$

In $ \triangle ADC,$

$ \angle D = 90^{\circ},$ using Pythagoras theorem.

$ \boxed{(AC)^{2} = (CD)^{2} + (AD)^{2}} $

$(4+r)^{2} = (4-r)^{2} + (4)^{2}$

$ \Rightarrow 16 + r^{2} + 8r = 16 + r^{2} – 8r + 16$

$ \Rightarrow 8r = -8r + 16 $

$\Rightarrow 8r + 8r = 16$

$ \Rightarrow 16r = 16$

$ \Rightarrow r = 1$

$ \therefore $ The radius of the third circle is $1\;\text{cm}.$

Correct Answer: D