Question

If ABCD is a cyclic quadrilateral with AB as the diameter of the circle and $\angle ACD={50}^{\circ },$ then $\angle BAD$ = ___.

• A. ${70}^{\circ }$
• B. ${80}^{\circ }$
• C. ${50}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is D

${40}^{\circ }$

Since an angle in a semicircle is a right angle, $\angle ACB={90}^{\circ }.$
But, $\angle BCD=\angle ACD+\angle ACB={50}^{\circ }+{90}^{\circ }={140}^{\circ }.$
Consider the cyclic quadrilateral ABCD. Since opposite angles are supplementary,
$\angle BCD+\angle DAB={180}^{\circ }.$
$⟹\angle BAD={180}^{\circ }-{140}^{\circ }={40}^{\circ }$