and 
.
and 
.
Equality holds when 
or 
be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
intersect 
at 
and 
are either parallel or coincide.
.
Where 
Y by Adventure10, Mango247
Let 
be an equilateral triangle. A point 
is chosen on 
and on arcs 
and 
of the circumcircle of , 
and 
are chosen respectively, so that 
is parallel to and 
is parallel to . Segments 
and intersect at point 
, while 
and intersect in point 
. Prove that the perimeters of the polygons 
and 
are the same.
be an equilateral triangle. A point 
is chosen on 
and on arcs 
and 
of the circumcircle of , 
and 
are chosen respectively, so that 
is parallel to and 
is parallel to . Segments 
and intersect at point 
, while 
and intersect in point 
. Prove that the perimeters of the polygons 
and 
are the same.
cut 
at 
and the circumcircle of 
again at 
respectively. 
and 
imply that 
are the midpoints of the arcs 
Thus, 
and 
and 
are kites 
and 
Hence, to prove that 
and the equilateral 
we have 
and 
Hence, 
as desired.
where 
is the intersection point of parallel line from 
and 
is the intersection point of parallel line from 
spiral homothety and Cyclic quadrilaterals help so much to find some other regular triangles.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/2/1/f/21f27b61fac12c932bacee1ded1f8873a242f76d.png)
is cyclic. Similarly, 
is cyclic.
Similarly, 
and 
,
so 
are parallel and all points lie on the circumcircle, we can say 
and 
.
,
