Question
Prove that the angle between the two tangent drawn from an external point to a circle is supplenentary to the angle subtended by the line segments joining the points of contact at the centre.
✓
Answer
Given $PA$ and $PB$ are the tangent drawn from a point P to a circle with centre $O$. Also, the line segments $OA$ and $OB$ are shown.
To prove: $\angle\text{APB}+\angle\text{AOB}=180^\circ$
Proof:
We know that the tangent is perpendicular to the radius through the point of contact.
$\therefore\text{PA}\perp\text{OA}\Rightarrow\angle\text{OAP}=90^\circ$
$\therefore\text{PB}\perp\text{OB}\Rightarrow\angle\text{OBP}=90^\circ$
$\therefore\angle\text{OBP}+\angle\text{OBP}=90^\circ+90^\circ=180^\circ\dots\text{(i)}$
But, we know that tha sum of all the angles of a quadrilateral is 360º.
$\therefore\angle\text{OAP}+\angle\text{APB}+\angle\text{AOB}+\angle\text{OBP}=360^\circ\dots\text{(ii)}$
From (i) and (ii), we get
$\angle\text{APB}+\angle\text{AOB}=180^\circ$
Hence proved.
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