In the given figure, ∠CED = ∠CAB.

(i) Show that :

△CED is similar to triangle CAB.

(ii) Find DE.

(i) In △ CED and △ CAB,

⇒ ∠CED = ∠CAB (Given)

⇒ ∠DCE = ∠ADB (Common angle)

∴ △ CED ~ △ CAB (By A.A. axiom)

Hence, proved that △CED is similar to triangle CAB.

(ii) We know that,

Corresponding sides of similar triangle are proportional.

$\therefore \dfrac{CE}{CA} = \dfrac{ED}{AB} \\[1em] \Rightarrow \dfrac{10}{15} = \dfrac{ED}{9} \\[1em] \Rightarrow ED = \dfrac{10}{15} \times 9 \\[1em] \Rightarrow ED = \dfrac{90}{15} = \text{6 cm.}$

Hence, DE = 6 cm.