Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Step-by-Step Solution
Key Concept: In quadrilateral $OAPB$, $\angle OAP = 90^\circ, \angle OBP = 90^\circ$. Sum of angles $= 360^\circ \Rightarrow \angle APB + \angle AOB = 180^\circ$.
Stepwise Solution:
Let $PA, PB$ be tangents from point $P$ to circle $C(O, r)$ touching at $A$ and $B$. [0.5 Mark]
$OA \perp PA \Rightarrow \angle OAP = 90^\circ$ and $OB \perp PB \Rightarrow \angle OBP = 90^\circ$. [0.5 Mark]
In quadrilateral $OAPB$:
$\angle APB + \angle OAP + \angle AOB + \angle OBP = 360^\circ$. [0.5 Mark]
$\angle APB + 90^\circ + \angle AOB + 90^\circ = 360^\circ \Rightarrow \angle APB + \angle AOB = 180^\circ$. Proved! [0.5 Mark]
Marking Scheme:
• Identifying $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$: 1.0 Mark
• Applying quadrilateral angle sum to get $\angle APB + \angle AOB = 180^\circ$: 1.0 Mark
Correct Answer: