In a circle, chords AB and CD intersect at E, such that AE=BE=4, and CE=8. Find DE.
", "hints_md": "This is an application of the [Power of a Point Theorem](https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem).", "hints_html": "This is an application of the Power of a Point Theorem.
", "editorial_md": "The Power of a Point Theorem states that for any point $E$ inside a circle, the product of the lengths of the segments of one chord passing through $E$ is equal to the product of the lengths of the segments of any other chord passing through $E$. \n\nThis can be expressed as:\n\n$AE\\cdot $$BE$ = $CE\\cdot $$DE$\n\nFurthermore, because $AE$ = $4$, $BE$ = $4$ and $CE$ = $8$, we can substitute these values into the theorem's equation:\n\n$4\\cdot4 = 8\\cdot $$DE$\n\nThis simplifies to: \n\n$16$ = $8 \\cdot DE$\n\nTo solve for $DE$, we divide both sides of the equation by $8$, thus:\n\n$DE$ = $\\frac{16}{8}$ = $2$.\n\n- botman", "editorial_html": "The Power of a Point Theorem states that for any point E inside a circle, the product of the lengths of the segments of one chord passing through E is equal to the product of the lengths of the segments of any other chord passing through E.
This can be expressed as:
AE\\cdotBE = CE\\cdotDE
Furthermore, because AE = 4, BE = 4 and CE = 8, we can substitute these values into the theorem’s equation:
4\\cdot4 = 8\\cdotDE
This simplifies to:
16 = 8 \\cdot DE
To solve for DE, we divide both sides of the equation by 8, thus:
DE = \\frac{16}{8} = 2.