
Equivalently, two radii are parallel to two sides of the parent triangle, as will be shown in examples 1 and 2 below.
Everything we’ve been considering so far seems to work smoothly for the -triangle.







Since the parent triangle is obtuse, its circumcenter lies outside, opposite the obtuse angle. Thus
is a (convex) quadrilateral.
If is the radius of the circumcircle of triangle
, then we have that
and
(see here). Since
is the circumcenter, we also have that
. Hence,
is a rhombus; in fact, a special rhombus with the property that the diagonal
is equal to the equal sides. Every rhombus is a parallelogram so we conclude that radius
is parallel to side
and radius
is parallel to side
.







Suppose that a triangle with circumcenter
has the property that radius
is parallel to side
and radius
is parallel to side
. Such a triangle is necessarily obtuse, so that the circumcenter
lies outside the triangle and we again have a quadrilateral
. Join
and let
. Since
is parallel to
, we have that
. Since
is the circumcenter, triangle
is isosceles and so
. The situation is shown below, with the circumcircle included:
Since is parallel to
, we have that
. Thus, triangle
is isosceles; moreover,
. By the inscribed angle theorem:
Thus, the interior angles of triangle are
and
.






Since the area satisfies , we have that
.
Normally, . And so we must have
. In turn, this implies
is not possible due to the fact that
and
. So we take
, which yields
. Using
we get
Consequently, .





No matter how the vertices are arranged, we’ll end up with a rhombus in which a diagonal is equal to the side-lengths. The interior angles of such a rhombus are . Now just consider the triangle that remains when
is removed.



Consider various arrangements of the rhombus , for example the one shown below:
Apply the inscribed angle theorem to see that it is not possible to have be a square.
Takeaway
In any triangle , let
be the side-lengths,
the orthocenter,
the circumcenter, and
the circumradius. Then the following statements are equivalent:
- the reflection of
over
is
- the area
satisfies
- the reflection of
over
is
and the reflection of
over
is
Task
- (Aufbau) In triangle
, let
be the side-lengths,
the circumradius,
the circumcenter,
the nine-point center, and
the orthocenter. PROVE that the following statements are equivalent:
- the reflection of
over
is
- the area
satisfies
- the circle with diameter
passes through vertices
and
- radius
is parallel to side
and radius
is parallel to side
- the reflection of
over
is
and the reflection of
over
is
is the geometric mean of
and
, and
is the geometric mean of
and
. (
and
are the feet of the altitudes from
and
, respectively.)