#### Question

Two equal chords AB and CD of a circle with centre O, intersect each other at point P inside the circle, prove that:

(i) AP = CP,

(ii) BP = DP

#### Solution

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 chord bisects it)

∴ MP = `1/2` AB = `1/2` CD = ND ...........(i)

In rt Δ OMB, OM^{2}= OB^{2} = MB^{2} ………………..(ii)

In rt Δ OND, ON^{2} = OD^{2} - ND^{2} ………………..(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