In the figure (1) given below, AP = 2PB and CP = 2PD.

(i) Prove that △ACP is similar to △BDP and AC ∥ BD.

(ii) If AC = 4.5 cm, calculate the length of BD.

(i) Given, AP = 2PB and CP = 2PD.

$\therefore \dfrac{AP}{PB} = \dfrac{2}{1} \text{ and } \dfrac{CP}{PD} = \dfrac{2}{1}.$

∠ APC = ∠ BPD [Vertically opposite angles]

So by SAS rule of similarity △ACP ~ △BDP.

Since, triangles are similar,

∴ ∠ CAP = ∠ PBD.

Since, these angles are alternate angles therefore, AC ∥ BD.

Hence, proved that △ACP ~ △BDP and AC ∥ BD.

(ii) Since triangles are similar. So,

$\dfrac{AP}{PB} = \dfrac{AC}{BD} \\[1em] \dfrac{AC}{BD} = \dfrac{2}{1} \\[1em] BD = \dfrac{AC}{2} \\[1em] BD = \dfrac{4.5}{2} \\[1em] BD = 2.25.$

Hence, BD = 2.25 cm.