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be a triangle, and let 
be a point on side 
. A line through intersects side 
at 
and ray 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle again at point 
distinct from point 
. The lines 
and 
intersect again at 
and 
respectively.
such that 
for all real number 
and 
be a scalene triangle with point 
and 
on the plane such that 
is an equilateral . Let intersect 
and 
at and respectively and intersect 
and 
at and respectively .
, point 
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds. is an isosceles triangle.
be an acute-angled triangle and the midpoint of . Let 
be a point on segment 
and 
its projection on . If 
and are the projections of on and then the interior angule bisectors of 
and 
are parallel.
, 
, 
, and 
, the angle bisector of is parallel to that of 
. Now if 
are such that 
, and 
, we require the angle bisectors of 
and to coincide. This, however is equivalent to 
. Denoting the first by and the second by , from sine law applied on 
, we compute:![\[\frac{\sin x}{AE}\cdot\frac{EM}{\cos x} = \frac{\sin\angle EAN}{NE}\cdot\frac{NE}{\sin\beta}=\frac{\sin\angle EAN}{\sin\beta} = \frac{\sin\angle EAP}{\sin\gamma} =\frac{\sin\angle EAP}{PE}\cdot \frac{PE}{\sin\gamma} =\frac{\sin y}{AE}\cdot\frac{EM}{\cos y} \]](http://latex.artofproblemsolving.com/3/d/3/3d3f0113ffc02f81b68ba64e71cff80f65311154.png)
Therefore 
, and we're done.
