Prove that if chords of congruent circles subtend equal angles at the center, then the chords are equal.

To prove: If chords of congruent circles subtend equal angles at the center, then the chords are equal.

Given:

Two congruent circles with centers O and O'.
Chords AB in the first circle and CD in the second circle.
The angles subtended by these chords at the respective centers are equal, i.e., ∠AOB = ∠CO'D.
Proof:

Since the circles are congruent, their radii are equal, i.e., OA = OB = O'C = O'D.
In circle 1, consider triangle AOB formed by the chord AB and radii OA and OB.
∠AOB is the angle subtended by chord AB at the center.
The triangle AOB is isosceles because OA = OB (radii of the circle).
Similarly, in circle 2, consider triangle CO'D formed by the chord CD and radii O'C and O'D.
∠CO'D is the angle subtended by chord CD at the center.
The triangle CO'D is isosceles because O'C = O'D (radii of the circle).
It is given that ∠AOB = ∠CO'D. Since the triangles AOB and CO'D have:
Two equal sides (radii of congruent circles),
An equal included angle (∠AOB = ∠CO'D), they are congruent by the SAS (Side-Angle-Side) congruence criterion.
By the congruence of triangles, the corresponding parts are equal, so AB = CD.
Hence, the chords AB and CD are equal if they subtend equal angles at the centers of congruent circles.

Q.E.D.