# 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

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by nishavini

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by nishavini

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The Brainliest Answer!

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....

: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 ...