Question

Given two equal chords AB and CD of a circle with center O, intersecting each other at point P.
Prove that:
(i) AP = CP
(ii) BP = DP 

Answer 1

In ΔOMP and ΔONP,
OP = OP                            ...( common sides )
∠OMP = ∠ONP                 ...( both are right angles )
OM = OM                          ...( side both the chords are equal, so the distance of the chords from the centre are also equal )
ΔOMP ≅ ΔONP                 ...( RHS congruence criterion )
⇒  MP = PN                       ...(c.p.c.t )  ....( a )

(i) Since AB = CD              ...( given )
⇒  AM = CN                     ...( drawn from the centre to the chord bisects the chord )
⇒  AM + MP = CN + NP  .....( from a )
⇒  AP = CP                       ....( b )

(ii) Since AB = CD
⇒ AP + BP = CP + DP
⇒  BP = DP                     ....( from  b )
Hence proved.