Hint: Use Pythagoras theorem on the various right angle triangles to get the desired result.

Complete step-by-step answer:

From the figure we can see that ABC is a right triangle, right angled at B and similarly $\vartriangle {\text{ABD}}$ is also a right triangle.

Using Pythagoras theorem in $\vartriangle {\text{ABD}}$

${\text{A}}{{\text{B}}^2} + B{D^2} = A{D^2} \\$

Now BC = 2BD as AD is a median (median divides a side in two equal parts)

${\text{A}}{{\text{B}}^2} + \dfrac{{B{C^2}}}{{{2^2}}} = A{D^2} \\$

So,

${\text{ 4A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2} = 4{\text{A}}{{\text{D}}^2}...........\left\{ 1 \right\} \\$

Now $\vartriangle {\text{BEC}}$ is also a right angled triangle so applying Pythagoras theorem

${\text{B}}{{\text{E}}^2} + {\text{B}}{{\text{C}}^2} = C{E^2} \\$

Now AB = 2BE as EC is a median (median divides a side in two equal parts)

$\dfrac{{{\text{A}}{{\text{B}}^2}}}{{{2^2}}} + B{C^2} = C{E^2} \\$

So,

${\text{ 4B}}{C^2} + A{B^2} = 4C{E^2}..............\left\{ 2 \right\} \\$

Adding equation (1) and (2), we get

${\text{5A}}{{\text{B}}^2} + 5B{C^2} = 4\left( {A{D^2} + C{E^2}} \right) \\$

But ${\text{ A}}{{\text{B}}^2} + B{C^2} = A{C^2} $, so now we get,

${\text{5A}}{{\text{C}}^2} = 4\left( {A{D^2} + C{E^2}} \right) \\$

$AC = 5$ cm and $AD = \dfrac{{3\sqrt 5 }}{2}$ cm

Substituting the values we get,

${\text{125 = 4}}\left( {\dfrac{{45}}{4} + C{E^2}} \right) \\$

On solving we get, $CE = 2\sqrt 5$ cm

Note: Always remember that a median divides a side in two equal parts so while solving such questions collaborate this concept with pythagoras theorem to reach a solution.