Question

Which of the following statement is correct?

• A. PM is parallel to AB.
• B. PM produced will not bisect the major arc AB.
• C. PM produced will bisect the major arc AB.
• D. none of the above

Answer 1

Answer: (C)

Given- O is the centre of a circle of which APB is an arc. P is the midpoint of the arc APB and M is the midpoint of the chord AB. PM is produced.
To find out- which of the given options is true.
Solution- We join OM, PA & PB. Now between $\Delta PAM$ & $\Delta PBM$, we have $PA=PB$(since the line, joining the midpoint of an arc and the midpoint of the chord contained by the arc, is perpendicular to the chord). Also $AM=BM$(given) and PM is common side.
$\therefore \Delta PAM\cong \Delta PBM$
$⟹\angle AMP=\angle BMP$.
But they are linear pair. $\therefore \angle AMP=\angle BMP={90}^o.......\left(i\right)$
Again OM meets AB at M which is the midpoint of AB.
$\therefore OM\perp AB⟹\angle OMA=\angle OMB={90}^o........\left(ii\right)$(because if a line through the centre of a circle bisects a chord of the same circle, then the line will be perpendicular to the same chord.)
So from (i) & (ii) $\angle BMP={90}^o=\angle OMB$.
$\therefore \angle BMP\&\angle OMB$ are linear pair.
$\therefore$ The points P, M & O are on the same straight line. Now we produce PO to meet the arc AQB at Q and join AQ & BQ. Now between $\Delta AQM$ & $\Delta BQM$, we have $AM=BM$, QM common side and $\angle OMA=\angle OMB={90}^o$ or $\angle QMA=\angle QMB={90}^o$ (from ii).
$\therefore$ By $SAS$ test $\Delta AQM=\Delta BQM⟹AQ=BQ.$
$\therefore arcARQ=arcBSQ.$ (since the line, joining the midpoint of an arc and the midpoint of the chord contained by the arc, is perpendicular to the chord.) So Q is the midpoint of the major arc AQB.
$\therefore$ PM produced will bisect the AB.
Ans- Option C.