(a) In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠ BAC +∠ ACD =90^°.

In \(\displaystyle \Delta \mathrm{OAC}\),

\[\mathrm{OA}=\mathrm{OC} \]

\[\Rightarrow \angle \mathrm{OCA}=\angle \mathrm{OAC} \]

Now, \(\displaystyle \angle \mathrm{OCD}=90^{\circ}\)

\[\begin{aligned} & \Rightarrow \angle \mathrm{OCA}+\angle \mathrm{ACD}=90^{\circ} \\ & \Rightarrow \angle \mathrm{OAC}+\angle \mathrm{ACD}=90^{\circ} \end{aligned} \]

or \(\displaystyle \angle \mathrm{BAC}+\angle \mathrm{ACD}=90^{\circ}\)

\(\displaystyle \Delta \mathrm{OAP} \cong \Delta \mathrm{OAS}\)

\[\therefore \angle 1=\angle 2 \]

Similarly, \(\displaystyle \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8\)

Also, \(\displaystyle \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ}\)

\[\begin{aligned} & \Rightarrow 2(\angle 1+\angle 4+\angle 5+\angle 8)=360^{\circ} \\ & \Rightarrow \angle \mathrm{AOB}+\angle \mathrm{COD}=180^{\circ} \end{aligned} \]

Similarly, \(\displaystyle \angle \mathrm{BOC}+\angle \mathrm{AOD}=180^{\circ}\)