Question

In the given figure, o is the centre of the circle. If chord AB=chord AC Op is perpendicular to AB and OQ is perpendicular to AC show that PB = QC

Answer 1

Given O is the centre of the circle.

$O P \perp A B$

where AB is a chord of the circle.

Perpendicular from centre bisects the chord.

⇒ AL = LB

Similarly, $O Q \perp A C$

where AC is a chord of the circle.

Perpendicular from centre bisects the chord.

⇒ AM = MC

But AB = AC

$\Rightarrow \frac{A B}{2}=\frac{A C}{2}$

⇒ LB = MC

Now, OP = OQ (Radii of same circle)

Equal chords are equidistant from centre.

⇒ OL = OM

Then, OP – OL = OQ – OM

⇒ LP = MQ

In triangles LPB and MQC,

LB = MC (side)

LP = MQ (side)

∠PLB = ∠QMC = 90° (angle)

Therefore, ΔLPB ≅ ΔMQC (by SAS congruence rule)

Corresponding parts of congruence triangles are congruent.

⇒ PB = QC

Hence proved.

To learn more...

1. 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.if angle PBA=30degree,show that seg PB is parallel to seg QC

https://brainly.in/question/7324621

2. O is the centre of the circle in which Seg AB and seg AC are congruent chords. Radius OP is perpendicular to chord AB and radius OQ is perpendicular to chord AC. If angle PBA=30°,show that seg PB is parallel to seg QC. Please solve immediately.

https://brainly.in/question/7177249

Answer 2

Answer:

Given O is the centre of the circle.

P parallel

where AB is a chord of the circle.

Perpendicular from centre bisects the chord.

⇒ AL = LB  

Similarly, OQ Parallel AC

where AC is a chord of the circle.

Perpendicular from centre bisects the chord.

⇒ AM = MC

But AB = AC

AB/2 = AC/2

⇒ LB = MC

Now, OP = OQ (Radii of same circle)

Equal chords are equidistant from centre.

⇒ OL = OM

Then, OP – OL = OQ – OM

⇒ LP = MQ

In triangles LPB and MQC,

LB = MC (side)

LP = MQ (side)

∠PLB = ∠QMC = 90° (angle)

Therefore, ΔLPB ≅ ΔMQC (by SAS congruence rule)  

Corresponding parts of congruence triangles are congruent.

⇒ PB = QC

Hence proved.

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