If two equal chords AB and CD intersects within the circle at E. Prove that arc DA = arc CB.

To prove that arc DA is equal to arc CB, we can make use of the Inscribed Angle Theorem. This theorem states that if an angle is inscribed in a circle, its measure is equal to half that of the intercepted arc.

Let's proceed with the proof:

1. Draw a circle with center O.
2. Draw two equal chords, AB and CD, intersecting at point E within the circle, such that AB and CD are not parallel.
3. Join points A and C with the center of the circle, point O.
4. Since AB and CD are equal chords, segments OA and OC bisect them respectively.
5. Therefore, segment OA is perpendicular to AB, and segment OC is perpendicular to CD. They bisect the chords into equal halves.
6. Let's consider angles OAD and OCB.
7. Angle OAD is an inscribed angle, and its measure is half the intercepted arc, arc DA.
8. Similarly, angle OCB is also an inscribed angle, and its measure is half the intercepted arc, arc CB.
9. Since both AB and CD are equal chords, segments OA and OC are equal.
10. Therefore, angles OAD and OCB have the same measure.
11. By the Transitive Property of Equality, angles OAD and OCB are equal.
12. According to the Inscribed Angle Theorem, angles inscribed in the same arc have the same measure.
13. Since angles OAD and OCB are inscribed angles with the same measure, they are both inscribed in arc CB.
14. Therefore, arc DA and arc CB are equal.

Thus, we have proven that arc DA is equal to arc CB, given that two equal chords AB and CD intersect within the circle at point E.