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. 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.

Invalid username
Login to AoPS