Question

Two equal chords AB and CD of a circle with center O, intersect each other at point P inside the circle.
Prove that: (i) AP = CP ; (ii) BP = DP

Answer 1

Drop OM and ON perpendicular on AB and CD.
Join OP, OB, and OD.

∴ OM and ON bisect AB and CD respectively.     ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
∴ MB = `1/2"AB" = 1/2"CD" = "ND"`....(i)

In right ΔOMB,
OM² = OB² - MB²                            ....(ii)
In right ΔOND,
ON² = OD² - ND²                           ....(iii)

From (i), (ii), and (iii),
OM = ON

In ΔOPM and ΔOPN,
∠OMP = ∠ONP          ....( both 90° )
OP = OP                     ....( common )
OM = ON                   ....( proved above )
By Right Angle-Hypotenuse-Side criterion of congruence,
∴ ΔOPM ≅ ΔOPN      ....( by RHS )

The corresponding parts of the congruent triangles are congruent.
∴ PM = PN                ....( c.p.c.t. )

Adding (i) to both sides,
MB + PM = ND + PN
⇒ BP = DP
Now, AB = CD
∴  AB - BP = CD - DP     ...( ∵ BP = DP )
⇒ AP = CP.