In the figure, $\angle BAD = 60^\circ$, $\;$ $\angle ABD = 70^\circ$, $\;$ $\angle BDC = 40^\circ$
- Prove that $AC$ is the diameter of the circle.
- Find $\angle ACB$.
- Find $\angle BCD$.
$ABCD$ is a cyclic quadrilateral.
$\therefore \;$ $\angle BAD + \angle BCD = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]
$\therefore \;$ $\angle BCD = 180^\circ - \angle BAD = 180^\circ - 60^\circ = 120^\circ$
In $\triangle BAD$, $\;$ $\angle BAD + \angle ABD + \angle ADB = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]
$\therefore \;$ $\angle ADB = 180^\circ - \left(\angle BAD + \angle ABD\right) = 180^\circ - \left(60^\circ + 70^\circ\right) = 50^\circ$
Now, $\angle ADC = \angle ADB + \angle BDC = 50^\circ + 40^\circ = 90^\circ$
But angle in a semicircle is a right angle.
$\therefore \;$ $\angle ADC$ is an angle in a semicircle.
$\implies$ $AC$ is the diameter of the circle.
Now, $\;$ $\angle ADB = \angle ACB$ $\;\;\;$ [angles of same segment are equal]
$\therefore \;$ $\angle ACB = 50^\circ$