# Difference between revisions of "Power of a Point Theorem/Introductory Problem 4"

## Problem

(ARML) Chords $AB$ and $CD$ of a given circle are perpendicular to each other and intersect at a right angle at $E$ . Given that $BE = 16, DE = 4,$ and $AD = 5$, find $CE$.

## Solution

$ADE$ is a right triangle with hypotenuse 5 and leg 4. Thus, by the Pythagorean Theorem, $AE = 3$ (or by just knowing your Pythagorean Triples). Applying the Power of a Point Theorem gives $AE\cdot BE = CE\cdot DE$, or $3\cdot 16 = x\cdot 4$. Solving gives $x = 12$.

Back to the Power of a Point Theorem.
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