Given,

AC = 8 cm,

AD = 3 cm and ∠ACB = ∠CDA

From figure,

∠CDA = 90°

∴ ∠ACB = ∠CDA = 90°

In right angled ∆ADC,

AC² = AD² + CD²

⇒ (8)² = (3)² + (CD)²

CD² = 64 – 9 = 55

⇒ CD = √55 cm

In ∆CDB and ADC.

∠BDC = ∠AD [each 90°]

∠DBC = ∠DCA [each equal to 90°-∠A]

∴ ∠CDB ∼ ∆ADC

Then,

⇒ CD² = AD × BD