Question

In the given figure, $O$ is the centre of a circle. If $AB$ and $AC$ are chords of the circle such that $AB=AC$ and $OP\perp AB,OQ\perp AC$, prove that $PB=QC$.

Answer 1

Given: $AB$ and $AC$ are two equal chords of a circle with centre $O$.

$OP\perp AB$ and $OQ\perp AC$.

To prove: $PB=QC$

Proof: $OP\perp AB$

$\Rightarrow AM=MB$ .... (perpendicular from centre bisects the chord)....(i)

Similarly, $AN=NC$....(ii)

But, $AB=AC$

$\Rightarrow \frac{AB}{2}=\frac{AC}{2}$

$\Rightarrow MB=NC$ ...(iii) ( From (i) and (ii) )

Also, $OP=OQ$ (Radii of the circle)

and $OM=ON$ (Equal chords are equidistant from the centre)

$\Rightarrow OP-OM=OQ-ON$

$\Rightarrow MP=NQ$ ....(iv) (From figure)

In $\Delta MPB$ and $\Delta NQC$, we have

$\angle PMB=\angle QNC$ (Each =${90}^{\circ }$)

$MB=NC$ ( From (iii) )

$MP=NQ$ ( From (iv) )

$\therefore \Delta PMB\cong \Delta QNC$ (SAS)

$\Rightarrow PB=QC$ (CPCT)