In the figure (2) given below, ∠ ADE = ∠ ACB.

(i) Prove that △s ABC and AED are similar.

(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.

(i) From the given figure,

∠ ADE = ∠ ACB (Given)
∠ A = ∠ A. (Common)

Hence, by AA rule of similarity △ABC ~ △AED.

(ii) Since triangles are similar,

$\therefore \dfrac{BC}{DE} = \dfrac{AB}{AE} = \dfrac{AC}{AD} \\[1em]$

From figure:
AD = AB - BD = 6 - 1 = 5 cm.

Consider,

$\dfrac{AB}{AE} = \dfrac{AC}{AD} \\[1em] \dfrac{6}{3} = \dfrac{AC}{5} \\[1em] AC = \dfrac{30}{3} \\[1em] AC = 10.$

Hence, the length of AC = 10 cm.