solution of regular polygons of problem no . 2

From the given relation , we have $A_1A_2\cdot A_1A_3 + A_1A_2\cdot A_1A_4 = A_1A_3 \cdot A_1A_4$…………..(1)
Also in the cyclic quadrilateral $A_1A_3A_4A_5$ , we have , by the $Ptolemy's Theorem$ ,
$A_4A_5\cdot A_1A_3$ +$ A_3A_4\cdot A_1A_5$ = $A_3A_5\cdot A_1A_4$ .
Since $A_1A_2A_3.........................A_n$ is a regular polygon , so we have $A_iA_j = A_jA_k$ for $1\leq i\leq j\leq k\leq n$.
Comparing (1) and (2) , we have $A-1A_4 = A_1A_5$ .
Since the two diagonals follows the relation $A-1A_4 = A_1A_5$ , then it is easy to say that there must be the same number pf vertices between $_1 $ and $A_4$ as between $A_1$ and $A_5$.
So , only possibility is $n = 7$

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