Two chords AB and CD of a circle intersect externally at E. If EC = 2 cm, EA = 3 cm and AB = 5 cm, find the length of CD.

We know that,

If two chords of a circle intersect externally, then the products of the length of segments are equal.

From figure,

EA × EB = EC × ED ….(1)

EB = EA + AB = 3 + 5 = 8 cm

Substituting values in equation (1) we get,

⇒ 3 × 8 = 2 × ED

⇒ 24 = 2 × ED

⇒ ED = $\dfrac{24}{2}$

⇒ ED = 12 cm.

⇒ CD = ED - EC = 12 - 2 = 10 cm

Hence, CD = 10 cm.