The correct option is A True
Given, In △ABC, ∠A=90 and AD⊥BC
In △ABC,
∠BAC+∠ABC+∠ACB=180
90+∠ABC+∠ACB=180
∠ABC+∠ACB=90 (I)
In △CAD,
∠CAD+∠ACD+∠ADC=180
∠CAD+∠ACD+90=180
∠CAD+∠ACD=90..(II)
Equating (I) and (II),
∠ABC+∠ACB=∠CAD+∠ACD
∠ABC=∠CAD...(III)
Similarly, ∠ACB=∠BAD...(IV)
Now, In △s, ABC and DAC
∠ABC=∠CAD ..(From III)
∠BAC=∠CDA (Each 90∘)
∠ACD=∠ACB (Common angle)
Thus, △ABC∼△DAC (AAA rule)