In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that:

(i) ∆APC and ∆BPD are similar.

(ii) If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.

(i) In ∆APC and ∆BPD, we have

∠APC = ∠BPD [Vertically opposite angles are equal]

∠ACP = ∠BDP [Alternate angles (as, AC || BD) are equal]

∴ ∆APC ~ ∆BPD [By A.A.]

Hence, proved that ∆APC ~ ∆BPD.

(ii) In similar triangles the ratio of corresponding sides are equal.

$\dfrac{\text{AC}}{\text{BD}} = \dfrac{\text{PA}}{\text{PB}}$ …………..(1) and,

$\dfrac{\text{AC}}{\text{BD}} = \dfrac{\text{PC}}{\text{PD}}$ ……………(2)

Solving (1) we get,

$\Rightarrow \dfrac{\text{AC}}{\text{BD}} = \dfrac{\text{PA}}{\text{PB}} \\[1em] \Rightarrow \dfrac{3.6}{2.4} = \dfrac{\text{PA}}{3.2} \\[1em] \Rightarrow \text{PA} = \dfrac{3.6}{2.4} \times 3.2 \\[1em] \Rightarrow \text{PA} = \dfrac{3}{2} \times 3.2 \\[1em] \Rightarrow \text{PA} = 4.8 \text{ cm}.$

Solving (2) we get,

$\Rightarrow \dfrac{\text{AC}}{\text{BD}} = \dfrac{\text{PC}}{\text{PD}} \\[1em] \Rightarrow \dfrac{3.6}{2.4} = \dfrac{\text{PC}}{4} \\[1em] \Rightarrow \text{PC} = \dfrac{3.6}{2.4} \times 4 \\[1em] \Rightarrow \text{PC} = \dfrac{3}{2} \times 4 \\[1em] \Rightarrow \text{PC} = 6 \text{ cm}.$

Hence, PA = 4.8 cm and PC = 6 cm.