Question 15 Marks
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer
Given A quad. ABCD circumscibes a circle with centre O.
To prove: $\angle\text{AOB}+\angle\text{COD}=180^\circ$
and $\angle\text{AOD}+\angle\text{BOC}=180^\circ$
Construction: Join OP, OQ, OR and OS.
Proof:
We know that tangents drawn from an external point to a circle subtend equal angles at the centre.
$\therefore\angle1=\angle2,\angle3=\angle4,\angle5=\angle6$ and $\angle7=\angle8$
and $\angle1+\angle2+\angle3+\angle4+\angle5+\angle6+\angle7+\angle8=360^\circ$
$\Rightarrow2(\angle2+\angle3)+2(\angle6+\angle7)=360^\circ$ and$2(\angle1+\angle8)+2(\angle4+\angle5)=360^\circ$
$\Rightarrow\angle2+\angle3+\angle6+\angle7=180^\circ$ and $\angle1+\angle8+\angle4+\angle5=180^\circ$
$\Rightarrow\angle\text{AOB}+\angle\text{COD}=180^\circ$ and $\angle\text{AOD}+\angle\text{BOC}=180^\circ$
Hence proved.
View full question & answer→Given A quad. ABCD circumscibes a circle with centre O.
To prove: $\angle\text{AOB}+\angle\text{COD}=180^\circ$
and $\angle\text{AOD}+\angle\text{BOC}=180^\circ$
Construction: Join OP, OQ, OR and OS.
Proof:
We know that tangents drawn from an external point to a circle subtend equal angles at the centre.
$\therefore\angle1=\angle2,\angle3=\angle4,\angle5=\angle6$ and $\angle7=\angle8$
and $\angle1+\angle2+\angle3+\angle4+\angle5+\angle6+\angle7+\angle8=360^\circ$
$\Rightarrow2(\angle2+\angle3)+2(\angle6+\angle7)=360^\circ$ and$2(\angle1+\angle8)+2(\angle4+\angle5)=360^\circ$
$\Rightarrow\angle2+\angle3+\angle6+\angle7=180^\circ$ and $\angle1+\angle8+\angle4+\angle5=180^\circ$
$\Rightarrow\angle\text{AOB}+\angle\text{COD}=180^\circ$ and $\angle\text{AOD}+\angle\text{BOC}=180^\circ$
Hence proved.
