In the figure (ii) given below, AB and CD are two intersecting chords of a circle. Name two triangles which are similar. Hence, calculate CP given that AP = 6 cm, PB = 4 cm, and CD = 14 cm (PC > PD).

In △APD and △CPB,

∠DAB = ∠DCB (∵ angles in same segment are equal.)

∠APD = ∠CPB (∵ vertically opposite angles are equal.)

△APD ~ △CPB. (By AA axiom)

Hence, proved that △APD ~ △CPB.

Chords AB and CD intersect each other at P.

Since △APD ~ △CPB, Hence, the ratio of corresponding sides are equal.

$\therefore \dfrac{AP}{CP} = \dfrac{PD}{PB}$

∴ AP × PB = CP × PD …..(i)

From figure,

CD = CP + PD

Let CP = x cm.

⇒ 14 = x + PD
⇒ PD = (x - 14) cm.

Putting values in eq (i)

⇒ 6 × 4 = x × (x - 14)
⇒ 24 = x² - 14x
⇒ x² - 14x - 24 = 0
⇒ x² - 12x - 2x - 24 = 0
⇒ x(x - 12) -2(x - 12) = 0
⇒ (x - 2)(x - 12) = 0
⇒ x - 2 = 0 or x - 12 = 0
⇒ x = 2 or x = 12

Since, given PC > PD so, CP = 12 cm.

Hence, the length of CP = 12 cm.