AE * EB = DE * EC
3 * 2 = 6 * EC
1 = EC
Alyssa Z.
asked 04/07/21Consider the following theorem.
If two chords intersect within a circle, then the product of the lengths of the segments (parts) of one chord is equal to the product of the lengths of the segments of the other chord.
O is the center of the circle.

| Given:AE = 3 |
| EB = 2 |
| DE = 6 |
Find:EC
AE * EB = DE * EC
3 * 2 = 6 * EC
1 = EC
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