In the given figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB=AC and OP⊥AB,OQ⊥AC, prove that PB=QC.
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Given: AB and AC are two equal chords of a circle with centre O.
OP⊥AB and OQ⊥AC.
To prove: PB=QC
Proof: OP⊥AB
⇒AM=MB .... (perpendicular from centre bisects the chord)....(i)
Similarly, AN=NC....(ii)
But, AB=AC
⇒AB2=AC2
⇒MB=NC ...(iii) ( From (i) and (ii) )
Also, OP=OQ (Radii of the circle)
and OM=ON (Equal chords are equidistant from the centre)
⇒OP−OM=OQ−ON
⇒MP=NQ ....(iv) (From figure)
In ΔMPB and ΔNQC, we have
∠PMB=∠QNC (Each =90∘)
MB=NC ( From (iii) )
MP=NQ ( From (iv) )
∴ΔPMB≅ΔQNC (SAS)
⇒PB=QC (CPCT)
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