Question

prove that the centre of a circle is on the right bisectors of each of its chords.


plz with full explanation not just the answer.. I need to solve my worksheet ​

Answer 1

The perpendicular bisector of any chord of any given circle must pass through the center of that circle. In the words of Euclid: From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line.

Answer 2

Answer:

Proved by congruence of triangles SAS property

Step-by-step explanation:

Here i am providing the detailed solution of this question

Let AB be the chord of circle and LM be the perpendicular bisector of that chord at R

To show Right bisectors of each of the chords passes through the center of circle

Let C be the point from where LM passes

To prove, C is the Centre of circle

Join the lines AC and BC

In triangle CAR and triangle CRB

AR=RB (CR ⊥AB)

∠CRA=∠CRB (90° each)

CR=CR (common)

ΔCAR ≅ΔCRB ( SAS criterion)

Now by CPCT( corresponding parts of congruent triangles)

CA= CB ( its is only possible when C is the center of circle)

Hence Proved that Right bisectors of each of the chords passes through the center of circle

#SPJ2