Question

O is the centre of a circle in which AB and AC are congruent chords. Radius OP is perpendicular to chord AB and radius OQ is perpendicular to chord AC. If $\angle PBA={30}^{\circ }$, show that PB is parallel to QC.

Answer 1

Given, $O$ is the center of a circle.

$AB=AC$

$OP\perp AB$

$OQ\perp AC$

$\angle PBA={30}^{\circ }$

To prove :- $BP\parallel QC$

Construction : Join $BC,$ $OC$ and $OB,$ as shown in the figure above.

Proof : $AB=AC$

We know that, angle opposite to equal sides of a triangle are equal.

$\therefore \angle ACB=\angle ABC$ __ (1)

$OC=OB$ $[\because$ radius]

$\angle OCB=\angle OBC$ __ (2)

Using (1), we have:

$\angle ACB=\angle ABC$

$\Rightarrow \angle ACO+\angle OCB=\angle ABO+\angle OBC$

Using (2), we can say:

$\Rightarrow \angle ACO=\angle ABO$ __ (3)

In $△OXC$ and $△OYB$

$\angle OXC=\angle OYB={90}^o[\because OQ\perp AC\&OP\perp AB]$

$\angle AOC=\angle ABO$ [using (3)]

$OC=OB$ [Radii]

$\therefore △OXC\cong △OYB$ [AAS]

$\angle QOC=\angle POB$ [CPCT] __ (4)

Now in $△QOC$ and $△POB$,

$OQ=OB[\because$ radius]

$\angle QOC=\angle POB$ (using (4))

$OC=OP(\because$ radius)

$\therefore △QOC\cong △POB$ [SAS]

$\therefore OQC=\angle OBP$ [CPCT]

These are alternate interior angles formed between $QC$ and $PB$.

Hence, $QC\parallel PB\left[Henceproved\right]$