Cyclic Quadrilateral

Relevant wiki: Cyclic Quadrilaterals

Include the central point $O$ of the circle. Since arc $DC$ is half the length of arc $BC$ , this shows that $\angle DOC = \theta$ and $\angle COB = 2\theta$ . Because point $O$ lies on diagonal $\overline{DB}$ , we can see that $\angle DOC + \angle COB = 180^{\circ}$ , where $\angle DOC = 60^{\circ}$ . The reason why this must holds follows that $\angle CAB = 2\angle COB$ . Then, $\angle COB = 120^{\circ}$ .

Being the central point of the circle, point $O$ forms two isosceles triangles $\Delta DOC$ and $\Delta COB$ . In that case,

• $\angle OCB = \angle OBC = 30^{\circ}$
• $\angle ODC = \angle OCD = 60^{\circ}$ , which implies $\Delta DOC$ is an equilateral triangle

Thus, since $\angle ACO = \angle ACB - \angle OCB = 15^{\circ}$ the answer is $\angle ACD = \angle DOC - \angle ACO = \boxed{45^{\circ}}$

which follows Thale's theorem . We can see that the sum of each pair of opposite angles of the cyclic quadrilateral is $180^{\circ}$ .

Note: Another way to approach this problem is to note that since $\overline{BD}$ is the diameter of the circle, one can note that two arc formed by $\overline{BD}$ have the same length. In that case, $\angle DCB = \angle DAB = 90^{\circ}$ . Thus, $\angle DCA = 45^{\circ}$ .

Note: It is also possible to solve directly, using the fact that the quadrilateral is cyclic. Then, you are to solve $\begin{array}{rl} \angle DAC + \angle CAD &= 75\\ \angle DAC + 60 &= \angle CAD + 45 \end{array}$ So $\angle DAC = 30^{\circ}$ and $\angle DCA = 45^{\circ}$ .

As O lies on BD, BD is the diameter of the circle. And as C lies on the periphery of the circle, the triangle BCD is a right triangle, therefor the angle at C is 90° (Thale's theorem) and the searched angle is 90°-45°= 45°

Keep in mind: The measurement of an inscribed (vertex on the circle) is always 1/2 the measurement of the intercepted arc. So...

because $<ACB=<ACD$ (Properties of a quadrilateral )so 45 degree