ISL 2002 G3
by Wolstenholme, Aug 4, 2014, 2:50 AM
The circle
has centre
, and
is a diameter of
. Let
be a point of
such that
. Let
be the midpoint of the arc
which does not contain
. The line through
parallel to
meets the line
at
. The perpendicular bisector of
meets
at
and at
. Prove that
is the incentre of the triangle 
Proof:
Since
we have that arcs
so
bisects
. Therefore, since
is the midpoint of arc
of circle
, it suffices to show that
. Now since segments
and
are perpendicular and bisect one another we have that quadrilateral
is a rhombus so
so it suffices to show that
so it suffices to show that
is a parallelogram. But
by definition and
since
is a diameter of
so
is in fact a parallelogram and we are done.




















Proof:
Since


















