(teaser) is it possible to have a polygon with number of diagonals twice the number of its sides?

A diagonal is a segment that connects two non-consecutive vertices in a polygon. The number of diagonals in a polygon that can be drawn from any vertex in a polygon is three less than the number of sides. To find the total number of diagonals in a polygon, multiply the number of diagonals per vertex (n - 3) by the number of vertices, n, and divide by 2

for example an Heptagon has 7 vertices

So n=7

(n-3) = (7-3) = 4

Multiply 4 by n , So 4*7 = 28

Now divide 28 by 2 = 14

SO heptagon has 14 diagonals

So Yes it is possible.

.bjbjk.b,j.l.

step-by-step explanation:

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)