Given : $\angle BAD = 65^{\circ}, \angle ABD = 70^{\circ}, \angle BDC$ $=45^{\circ}$ $\because ABCD$ is a cyclic quadrilateral. In $\triangle ABD$, $\angle BDA + \angle DAB + \angle ABD = 180^{\circ}$ (By using sum property of $\Delta^0$ ) $\therefore \angle BDA = 180^{\circ} - (65^{\circ} + 70^{\circ})$ $=180^{\circ} - 135^{\circ} = 45^{\circ}$ Now from $\triangle A C D$, $\angle ADC = \angle ADB + \angle BDC$ $=45^{\circ} + 45^{\circ}$ $( \because \angle BDA = \angle ADB = 45^{\circ} )$ $=90^{\circ}$ Hence, $\angle D$ makes right angle belongs in semi-circle therefore $A C$ is a diameter of the circle.