The given figure shows two congruent circles with centres P and Q. R is mid-point of PQ and ABRCD is a straight line.

Prove that : AB = CD.

Given: Two congruent circles with centres P and Q. Point R is the midpoint of PQ and ABRCD is a straight line.

To proof: AB = CD

Construction: Draw PM ⊥ AB and QO ⊥ CD.

Proof: In triangles Δ MPR and Δ OQR:

∠MRP = ∠ORQ (Vertically opposite angles)

∠PMR = ∠QOR = 90° (Since PM ⊥ AB and QO ⊥ CD)

RP = RQ (Since R is the midpoint of PQ)

So, by ASA congruency criterion:

Δ MPR ≅ Δ OQR

By corresponding parts of congruent triangles,

⇒ PM = QO

Now, in congruent circles, if the perpendicular distances of two chords from the centres are equal, then the chords are also equal.

⇒ AB = CD

Hence, AB = CD.