In the given figure, ∠A = ∠CED, prove that ∆CAB ∼ ∆CED. Also, find the value of x.

Comparing ΔCAB and ΔCED,

∠CAB = ∠CED                [Given]

∠ACB = ∠ECD                [Common]

$\therefore \triangle C A B \sim \triangle C E D$

$\Rightarrow \frac{C A}{C E}=\frac{A B}{E D}$ [In similar triangles, corresponding sides are in the same proportion]

$\Rightarrow \frac{15 \mathrm{~cm}}{10 \mathrm{~cm}}=\frac{9 \mathrm{~cm}}{x}$

$\Rightarrow x=\frac{9 \times 10}{15} \mathrm{~cm}=6 \mathrm{~cm}$