Answer:

The proof in explained below.

Step-by-step explanation:

Given, O is the centre of a circle in which segment AB and segment AC are congruent chords. radius OP is perpendicular to chord AB and radius OQ is perpendicular to chord AC.

We have to prove that segment PB is parallel to segment QC

Lets do first construction join OC, OB and BC so that OC and OB becomes the radius of circle.

We have to prove that PC and BQ makes the straight line so that we can prove the alternate opposite angles that are ∠QCP and ∠CPB equal in order to prove PB||QC.

AC=AB ⇒ ∠ACB=∠ABC

OC=OB ⇒ ∠OCB=∠OBC

⇒∠ACB-∠OCB=∠ABC-∠OBC ⇒ ∠5=∠6

In ΔOXC and ΔOYB

∠1=∠2 (∵each 90°)

∠5=∠6 (proved above)

OC=OB (∵radii of same circle)

By AAS rule, ΔOXC≅ΔOYB

By CPCT, ∠2=∠1 i.e vertically opposite angles are equal.

Hence, CP and BQ are striaght lines.

In ΔOCQ and ΔOPB

OQ = OB (∵radii of same circle)

OC = OP (∵radii of same circle)

∠1=∠2 (proved above)

Hence, by SAS rule, ΔOCQ≅ΔOPB

By CPCT, ∠OCQ=∠OPB

But these are alternate angles

⇒ PB||QC