If $PA$ and $PB$ are two tangents drawn from an external point $P$ to a circle with centre $O$, prove that $OP$ is the perpendicular bisector of chord $AB$.
Step-by-Step Solution
Key Concept: $PA = PB$ and $OA = OB \Rightarrow P$ and $O$ are equidistant from $A$ and $B \Rightarrow OP$ is perpendicular bisector of $AB$.
$\Delta OAP \cong \Delta OBP$ (RHS congruence criterion) $\Rightarrow \angle APO = \angle BPO$. [1.0 Mark]
In $\Delta APM$ and $\Delta BPM$, $PA = PB, \angle APM = \angle BPM, PM = PM \Rightarrow \Delta APM \cong \Delta BPM \Rightarrow AM = MB$ and $\angle AMP = 90^\circ$. $OP \perp AB$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Proving $\Delta OAP \cong \Delta OBP$: 1.0 Mark
Deducing $OP$ is perpendicular bisector of $AB$: 1.0 Mark
Correct Answer: