Ptolemy's Theorem and Cyclic Quadrilateral

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SUMMARY

The discussion focuses on solving a problem involving Ptolemy's Theorem in a cyclic quadrilateral ABCD, where the diagonals intersect at point E. Given the lengths AB=5, BC=10, BE=7, and CD=6, the objective is to find the length of CE. The key to solving this problem lies in applying Ptolemy's Theorem, which states that the product of the diagonals equals the sum of the products of opposite sides. Additionally, the similarity of triangles ΔDEC and ΔAEB is established through the AAA criterion, utilizing the theorem regarding angles subtended by the same arc.

PREREQUISITES
  • Ptolemy's Theorem
  • Properties of cyclic quadrilaterals
  • Triangle similarity (AAA criterion)
  • Theorem on angles subtended by the same arc
NEXT STEPS
  • Study the applications of Ptolemy's Theorem in various geometric problems
  • Explore the properties of cyclic quadrilaterals in depth
  • Learn about triangle similarity and its proofs
  • Investigate the theorem regarding angles subtended by the same arc
USEFUL FOR

Students studying geometry, particularly those focused on cyclic quadrilaterals and Ptolemy's Theorem, as well as educators looking for problem-solving strategies in geometric contexts.

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Homework Statement


In cyclic quadrilateral ABCD with diagonals intersecting at E, we have AB=5, BC=10, BE=7, and CD=6. Find CE.

Homework Equations


Ptolemy's Theorem: The product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of opposite sides.

The Attempt at a Solution


I drew the picture and that's as far as I got.
 

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brisk11228 said:

Homework Statement


In cyclic quadrilateral ABCD with diagonals intersecting at E, we have AB=5, BC=10, BE=7, and CD=6. Find CE.

Homework Equations


Ptolemy's Theorem: The product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of opposite sides.

The Attempt at a Solution


I drew the picture and that's as far as I got.

Hint : First prove that ΔDEC is similar to ΔAEB by axiom AAA. To prove them similar , you have to apply theorem related to angles subtended by same segment or arc on circumference. Do you know what does that theorem state ?
 

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