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