If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Sure, here is the proof:

Let's consider a circle with center O. Let AB and CD be two equal chords of the circle that intersect at point E within the circle. We need to prove that AE = CE and BE = DE.

Step 1: Draw radii OA and OC to the endpoints of the chords.

Step 2: Since AB = CD (given) and OA = OC (radii of the same circle), we have ΔOAB ≅ ΔOCD by the Side-Side-Side (SSS) criterion of congruence.

Step 3: From the congruence of the triangles, we have ∠OAB = ∠OCD and ∠OBA = ∠ODC (corresponding parts of congruent triangles are equal).

Step 4: Now, consider triangles AOE and COE. We have ∠OAE = ∠OCE (from step 3) and ∠AOE = ∠COE (vertically opposite angles).

Step 5: Therefore, by the Angle-Angle (AA) criterion, we have ΔAOE ≅ ΔCOE.

Step 6: From the congruence of these triangles, we have AE = CE (corresponding parts of congruent triangles are equal).

Step 7: Similarly, by considering triangles BOE and DOE, we can prove that BE = DE.

Therefore, the segments of one chord are equal to the corresponding segments of the other chord.