Given: circle k(O), O∈ AB ,CD ⊥ AB Prove: △ADC∼△ACB

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Step-by-step explanation:

Given : In a circle O,

AB is a diameter and CD⊥AB,

To Prove :

ΔADC ~ ΔACB

Solution :

In ΔADC and ΔACB,

m∠ADC = 90° [Given]

m∠ACB = 90° [Angle subtended by the diameter = 90°]

m∠ADC ≅ m∠ACB ≅ 90°

∠A ≅ ∠A [Reflexive property]

Therefore, ΔADC ~ ΔACB [By AA postulate of similarity]