Diagonals of regular heptagon

Question

If $a,b$ are the length of unequal diagonals of a regular heptagon with side $c$, then Prove that $c(a+b)=ab$ .

Answer 1

Let $A_1, A_2, A_3, A_4, A_5, A_6, A_7$ be the vertices of the regular heptagon (is it seven-gon? I hope I got the number of edges right :)), listed in cyclic order. Let $k$ be the circle circumscribed around the heptagon. Triangle $A_1A_2A_5$ is isosceles with $A_1A_5 = A_2A_5 = a$ (follows by regularity and hence symmetry of the heptagon). Triangle $A_1A_3A_5$ is also isosceles with $A_1A_3 = A_3A_5 = b$ again by regularity and hence symmetry of the heptagon. By regularity of the heptagon, $A_1A_2 = A_2A_3 = c$. Finally, the quadrilateral $A_1A_2A_3A_5$ is inscribed in the circle $k$ and therefore it satisfies Ptolemy's theorem, according to (one direction of) which $$A_1A_2 \cdot A_3A_5 + A_2A_3 \cdot A_1A_5 = A_1A_3 \cdot A_2A_5$$ or written in $a, b, c$ notations $$cb + c a = b a$$ which is exactly the relation $c(a + b) = ab$.