In ΔABC angle ∠C = 90° and M and n are mid points of the sides AB and AC respectively. CM and BN intersect each other at D and ∠BDC = 90°. If BC = 12 cm, then the length of BN is ?

6\(\sqrt {6}\)

BN and CM are the medians 

So, D divide BN in the ratio 2 : 1

⇒ BD : DN = 2 : 1

Let BD = 2x , DN = x

⇒ BN = 3x

In ΔCNB, CD is perpendicular to BN

(BC)² = (BD) × (BN)

(12)² = 2x × 3x

144 = 6x² ⇒ x² = \(\frac{144}{6}\)

x = 2\(\sqrt {6}\) cm

3x = 3(2\(\sqrt {6}\) )cm = 6\(\sqrt {6}\) cm