Given : In circle C (O, r) equal chords AB and CD intersects at P.
To Prove: AP = CP and BP = DP.
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOPM and ΔONP,
∠OMP = ∠ONP [∵ Each 90]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
PM = PN …(1) [∵ CPCT]
And AB = CD …(2) [∵ Given]
⇒ (1/2)AB = (1/2)CD
⇒ AM = CN …(3)
Adding the equations (1) and (3), we have
AM + PM = CN + PM
⇒ AP = CP …(4)
Subtracting equation (4) from (2), we have
AB – AP = CD – CP
⇒ PB = PD