Answer:
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]
