In the given figure, \(AB\) is a dimeter of the circle with centre \(O\). \(AQ\), \(BP\) and \(PQ\) are tangents to the circle. Prove that \(\angle POQ = 90^\circ\).
Proof:
Draw: Point \(O\) meets \(PQ\) at \(R\).
In \(\Delta OBP\) and \(\Delta OPR\),
\(BP =\) ()
\(OP =\) ()
\(OB =\) ()
Thus, \(\Delta OBP \cong\) (by ).
Hence, \(\angle BOP =\) (by CPCT).
Similarly, \(\Delta AOQ \cong \)
\(\angle AOQ = \angle ROQ\) (by )
Since \(AOB\) is a straight line,
\(\angle AOB = 180^\circ\)
\(= 180^\circ\)
\(= 180^\circ\)
\(= 180^\circ\)
\(\angle QOR + \angle ROP = \)
\(\angle POQ = 90^\circ\)
Hence proved.
Answer variants:
\(ASA\) congruence
Tangents from same external point
\(SAS\) congruence
\(\Delta ROQ\)
\(OP\)
CPCT
\(PR\)
\(SSS\) congruence
\(\angle QOR + \angle QOR + \angle ROP + \angle ROP\)
\(2 \angle QOR + 2 \angle ROP\)
\(\angle POR\)
Radii
\(90^\circ\)
\(\Delta OPR\)
\(\angle AOQ + \angle QOR + \angle ROP + \angle BOP\)
Common side
\(OR\)