Question

IN THE GIVEN FIGURE O IS THE CENTRE OF CIRCLE,OL IS PERPENDICULAR TO AB AND OM IS PERPENDICULAR TO CB,IF ANGLE OAB=OCB,THEN PROVE THAT AB=CB

Answer 1

Here AB and CB is the chord.
And as we know that Perpendicular line from centre of circle bisect the cord.
So in chord AB, AL = LB
Similarly in chord CB, CM = MB
Now in triangle AOL and COM, we have
::Angle CLA = Angle AMC = 90⁰
::OA = OC                  (BOTH ARE RADIUS)
::Angle LOA = Angle MOC          (Vertically opposite angles)

Therefore, triangle AOL is congruent to triangle COM by AAS criterion of congruency
=> AL = MC                                      (BY C.P.C.T)
=>2AL = 2MC                                    (Multiplying both the sides by 2)
=>AB = BC                                        (2AL = AB and 2MC = BC)

Hence both the chords are equal....


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Answer 2

given
:o is the center
:ol and om are perpendicular to ab and cb respectively
:<oab=<ocb
to prove
:ab=cb
PROOF
consider tri.AOL and COM
<l=<m=90
<loa=<moc (vert.opp.ang.)
oa=oc  (radii)
hence,triangle AOL is congruent to tri. COM by aas congruence criteria
so,  AL=CM (CPCT)
but, L and M are midpoints of ab and cb as OL and OM are the perpendiculars from the center and the perpendiculars from the center to the chord bisect the chord.
Thus, 2AL=2CM
AL=LB & CM=MB
so, AB=CB
HENCE THE PROOF ...