Question

of a circle which circumscribes
4. In the given figure, O is the centre of a circle. If AB and CD
equal chords of the circle, OP 1 AB and OQ 1 CD,prove that PB=QC​

Answer 1

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.

PLZ MARK AS BRIANLIEST,FLW ME AND THX FOR THE SUPERB QUESTION

Answer 2

Answer:

hope so it will helpful to you