In the figure (i) given below, chords AB and CD of a circle intersect at E.

(i) Prove that triangles ADE and CBE are similar.

(ii) Given DC = 12 cm, DE = 4 cm and AE = 16 cm, calculate the length of BE.

(i) In △CBE and △ADE,

∠B = ∠D (∵ angles in same segment are equal.)

∠BEC = ∠DEA (∵ vertically opposite angles are equal.)

△CBE ~ △ADE. (By AA axiom)

Hence, proved that △CBE ~ △ADE.

(ii) Given, DC = 12 cm.

From figure,

⇒ DC = DE + EC
⇒ 12 = 4 + EC
⇒ EC = 12 - 4
⇒ EC = 8 cm.

Chords AB and CD intersect each other at E.

Considering △BEC and △AED,

∠BEC = ∠DEA (∵ vertically opposite angles are equal.)

∠CBE = ∠EDA (∵ both angles are subtended on circle by arc AC and angles in same segment are equal.)

Hence, △BEC ~ △AED.

Since triangles are similar hence the ratio of the corresponding sides are equal.

$\therefore \dfrac{BE}{DE} = \dfrac{EC}{EA} \\[1em] \Rightarrow BE \times EA = EC \times DE \\[1em] BE \times 16 = 8 \times 4 \\[1em] BE = \dfrac{32}{16} \\[1em] BE = 2.$

Hence, the length of BE = 2 cm.