In $\triangle$ ABC, $\angle C = 90^\circ$. M and N are the mid-points of sides AB and AC,respectively. CM and BN intersect each other at D and $\angle BDC = 90^\circ$. If BC = 8 cm, then the length of BN is:

$4\sqrt{6}$ cm

As BN and CM are medians of triangle ABC, D is centroid of the triangle.

So, D will divide BN in the ratio 2 : 1

= BD : DN = 2 : 1

Let BD = 2x and DN = x

= BN = 3x

Also in right angled triangle CNB, CD perpendicular to BN

= \( {BC }^{ 2} \) = BD x BN

= \( {8}^{ 2} \) =  2x x 3x

= 64 = \( {6x }^{ 2} \)

= x = \(\frac{4√6}{3}\)

= BN = 3x = 3 x (\(\frac{4√6}{3}\)) = 4√6

Therefore, BD is 4√6 cm.