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In the figure if angleACB = angleCDA, AC = 8 cm and AD= 3 cm, then BD is
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Answer:

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,

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