ABCD is a cyclic quadrilateral, AB is diameter of the circle. If angle ∠ACD = 45^o, then what is the value of ∠BAD ?

45^o

According to the concept,

\(\angle\)ADB = \(\angle\)ACB = \({90}^\circ\)

Now,

\(\angle\)BCD

= \(\angle\)ACD + \(\angle\)ACB

= 45 + 90

= \({135}^\circ\)3

According to the concept,

= \(\angle\)BAD + \(\angle\)BCD = \({180}^\circ\)

= \(\angle\)BAD + \({135}^\circ\) = \({180}^\circ\)

= \(\angle\)BAD = \({180}^\circ\) - \({135}^\circ\)

= \(\angle\)BAD = \({45}^\circ\)

Therefore, \(\angle\)BAD is \({45}^\circ\).