# 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>. | + | ([[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 == | ||

− | <math> ADE </math> is a right triangle with hypotenuse 5 and leg 4. Thus, by the [[Pythagorean Theorem]], <math> AE = 3 </math> (or by just knowing your [[Pythagorean Triple]]s). Applying the Power of a Point Theorem gives <math> AE\cdot BE = CE\cdot DE </math> or <math> 3\cdot 16 = x\cdot 4 </math>. Solving gives <math> x = 12 </math>. | + | <asy> |

+ | import olympiad; | ||

+ | pair E = (0,0); | ||

+ | pair A = (3,0); | ||

+ | pair D = (0,4); | ||

+ | pair B = (-16,0); | ||

+ | pair C = (0,-12); | ||

+ | dot("E", E, SW); | ||

+ | dot("A", A, dir(0)); | ||

+ | dot("D", D, N); | ||

+ | dot("B", B, W); | ||

+ | dot("C", C, SE); | ||

+ | draw(A--D--E--cycle); | ||

+ | draw(B--E); | ||

+ | draw(C--E); | ||

+ | label("5", A--D, NE); | ||

+ | label("3", E--A, S); | ||

+ | label("4", D--E, W); | ||

+ | label("16", B--E, N); | ||

+ | label("x", C--E, W); | ||

+ | markscalefactor=0.1pt; | ||

+ | draw(rightanglemark(D, E, A)); | ||

+ | draw(circumcircle(B, D, A)); | ||

+ | </asy> | ||

+ | <math> ADE </math> is a right triangle with hypotenuse 5 and leg 4. Thus, by the [[Pythagorean Theorem]], <math> AE = 3 </math> (or by just knowing your [[Pythagorean Triple]]s). Applying the Power of a Point Theorem gives <math> AE\cdot BE = CE\cdot DE </math>, or <math> 3\cdot 16 = x\cdot 4 </math>. Solving gives <math> x = 12 </math>. | ||

''Back to the [[Power of a Point Theorem]].'' | ''Back to the [[Power of a Point Theorem]].'' |

## Latest revision as of 02:14, 26 November 2020

## Problem

(ARML) Chords and of a given circle are perpendicular to each other and intersect at a right angle at . Given that and , find .

## Solution

is a right triangle with hypotenuse 5 and leg 4. Thus, by the Pythagorean Theorem, (or by just knowing your Pythagorean Triples). Applying the Power of a Point Theorem gives , or . Solving gives .

*Back to the Power of a Point Theorem.*