ABCD is a cyclic quadrilateral such that AB is the diameter of the circle and ∠ADC = 145°, then what is the measure of ∠BAC ?

55°

\(\angle\)ADC + \(\angle\)ABC = \({180}^\circ\)    (Since the sum of opposite angles of a quadrilateral is \({180}^\circ\))

= \(\angle\)ABC = \({180}^\circ\) - \({145}^\circ\)

= \(\angle\)ABC = \({35}^\circ\)

Also, \(\angle\)ACB = \({90}^\circ\)    (since the angle subtended by diameter at the circumference of the circle is \({90}^\circ\))

In \(\Delta \)ABC, we have

\(\angle\)ACB = \({90}^\circ\) and \(\angle\)ABC = \({35}^\circ\)

So, \(\angle\)CAB = \({180}^\circ\) - (\(\angle\)ACB + \(\angle\)ABC)

= \({180}^\circ\) - (\({90}^\circ\) + \({35}^\circ\))

= \({180}^\circ\) - \({125}^\circ\)

= \({55}^\circ\)

Therefore, \(\angle\)CAB is \({55}^\circ\).