CBSE 2025 · Region 1 · Set 1 · Q26 · 3 marks
In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $\displaystyle \angle \mathrm{BAC}+\angle \mathrm{ACD}=90^{\circ}$.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $\displaystyle \angle \mathrm{BAC}+\angle \mathrm{ACD}=90^{\circ}$.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Marking-scheme solution
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, \(\displaystyle \angle \mathrm{BOC}+\angle \mathrm{AOD}=180^{\circ}\)
CirclesTangent Properties and ProofsApplyshort_answerhard
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.