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

[Albert1]

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 :$(a+b)^2(a-b)=ab^2$

 

[lfdahl]

Hi, Albert!

Spoiler

A hint is requested

 

[Albert1]

Hi, Albert!

Spoiler

A hint is requested

hint:

Spoiler

Using “Ptolemy’s Theorem” on quadrilaterals ABDG and BDEG

 

[lfdahl]

My attempt with the help from Alberts hint:

Spoiler

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:

$(a+b)^2(a-b) = ab^2$.