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

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== Problem ==
 
== Problem ==
([[ARML]]) Chords <math> AB </math> and <math> CD </math> of a given circle are perpendicular to each other and intersect at a right angle.  Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>.
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([[ARML]]) Chords <math> AB </math> and <math> CD </math> of a given circle are perpendicular to each other and intersect at a right angle at <math>E</math> .  Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>.
  
 
== Solution ==
 
== Solution ==

Revision as of 01:25, 20 February 2020

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.