Question : ABCD is a cyclic quadrilateral, AB is the diameter of the circle. If angle $\angle ACD=45^{\circ}$, then what is the value of $\angle BAD$?
Option 1: $90^{\circ}$
Option 2: $45^{\circ}$
Option 3: $135^{\circ}$
Option 4: $35^{\circ}$
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Correct Answer: $45^{\circ}$
Solution :
Since, an angle subtended by the diameter of a circle at the circumference = $90^{\circ}$
⇒ $\angle$ ACB = $90^{\circ}$
$\angle$ BCD = $\angle$ ACD + $\angle$ ACB = $45^{\circ} + 90^{\circ}=135^{\circ}$
Since the opposite angles of a cyclic quadrilateral are supplementary,
$\angle$ BCD + $\angle$ BAD = $180^{\circ}$
$\therefore \angle$ BAD = $180^{\circ} - 135^{\circ}=45^{\circ}$
Hence, the correct answer is $45^{\circ}$.
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