Question 177776
{{{ drawing( 300, 300, -12, 12, -12, 12, grid(1),
circle(10,0,.3),
circle(2,4,.3),
circle(-8,-6,.3),
circle(6,-8,.3),
green(line(10,0,6,-8)),
green(line(2,4,-8,-6)),
green(line(6,-8,-8,-6)),
green(line(2,4,10,0))) }}} 
Looks like a kite.
If it is then,
{{{DE=DG}}}
{{{FE=FG}}}
Use the distance formula,
{{{D^2=(x[1]-x[2])^2+(y[1]-y[2])^2}}}
For DE,
{{{D^2=(10-2)^2+(0-4)^2}}}
{{{D^2=64+16}}}
{{{highlight( D=sqrt(80)))}}}
For DG,
{{{D^2=(10-6)^2+(0-(-8))^2}}}
{{{D^2=16+64}}}
{{{highlight( D=sqrt(80))}}}
.
.
.
For FE,
{{{D^2=(-8-2)^2+(-6-4)^2}}}
{{{D^2=100+100}}}
{{{highlight( D=sqrt(200))}}}
For FG,
{{{D^2=(-8-6)^2+(-6-(-8))^2}}}
{{{D^2=196+4}}}
{{{highlight( D=sqrt(200))}}}
.
.
.
We have have proved that the classification is a kite because,
{{{DE=DG}}}
{{{FE=FG}}}