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

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SUMMARY

The discussion focuses on proving the heptagon diagonal equation \((a+b)^2(a-b)=ab^2\) where the side lengths of heptagon \(ABCDEFG\) are all equal to 1. The diagonals \(\overline{AD}\) and \(\overline{BG}\) are defined as \(a\) and \(b\) respectively, with the condition \(a > b\). The proof utilizes geometric properties and relationships inherent in regular heptagons.

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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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