How to Prove the Heptagon Diagonal Equation (a+b)^2(a-b)=ab^2?

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Discussion Overview

The discussion revolves around proving the equation \((a+b)^2(a-b)=ab^2\) related to the diagonals of a heptagon with equal side lengths. The focus is on the mathematical reasoning and proof techniques involved in this geometric problem.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the problem by defining the side lengths of a heptagon as equal to 1 and specifies the diagonals \(\overline{AD}=a\) and \(\overline{BG}=b\) with the condition \(a>b\).
  • Another participant greets the first, indicating a social interaction rather than contributing to the mathematical discussion.
  • A subsequent post hints at a potential approach to the proof, although the content of the hint is not provided.
  • A later reply indicates an attempt to solve the problem using the hint, but does not detail the method or results of that attempt.

Areas of Agreement / Disagreement

The discussion does not show clear agreement or disagreement on the proof itself, as the mathematical content is limited and primarily consists of social exchanges and hints.

Contextual Notes

The discussion lacks detailed mathematical steps or assumptions that may be necessary for a complete proof. The hint provided is not elaborated upon, leaving the proof approach ambiguous.

Who May Find This Useful

Participants interested in geometric proofs, particularly those involving polygons and diagonal relationships, may find this discussion relevant.

Albert1
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The measurement of each side length of a heptagon $ABCDEFG$ is equal to 1,
the diagonal of $\overline{AD}=a,$ and the diagonal of $\overline{BG}=b \,\,(a>b)$
prove :$(a+b)^2(a-b)=ab^2$
 
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Hi, Albert!

A hint is requested :o
 
lfdahl said:
Hi, Albert!

A hint is requested :o
hint:
Using “Ptolemy’s Theorem” on quadrilaterals ABDG and BDEG
 
My attempt with the help from Alberts hint:
View attachment 6459

From the figure, we have:

Ptolemys Theorem applied on quadrilateral ABDG: $a+b = ab$.

and applied on quadrilateral BDEG: $a^2 = b+b^2 \Rightarrow (a+b)(a-b) = b$.

Multiplying the left hand sides and the right hand sides yields the result:

$(a+b)^2(a-b) = ab^2$.
 

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  • Ptolemy´s Theorem applied.PNG
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