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

m (→‎Problem)
 
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== Solution ==
 
== Solution ==
 +
<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>.
 
<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 01:14, 26 November 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

[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] $ADE$ 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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