The Dual of a Theorem relative to the Orthocenter of a Triangle Professor Ion Patrascu, National College "BuzeĹ&#x;ti Brothers" Craiova - Romania Professor Florentin Smarandache, University of New Mexico, Gallup, USA
In [1] we introduced the notion of Bobillier transversal relative to a point the plane of a triangle
in
; we use this notion in what follows.
We transform by duality with respect to a circle
the following theorem
relative to the orthocenter of a triangle. Theorem 1. If
is a nonisosceles triangle,
its orthocenter, and
are cevians of a triangle concurrent at point and
different from
are the intersections of the perpendiculars taken from
cevians respectively, with
,
, then the points
Proof. We note with
,
on given
are collinear. , see
Figure 1. According to Ceva’s theorem, trigonometric form, we have the relation:
We notice that:
Because:
as angles of perpendicular sides, it follows that
1
Therewith
We thus get that:
Analogously, we find that: Figure 1.
Applying the reciprocal of Menelaus' P
theorem, we find, in view of (1), that:
This shows that
are collinear.
Note. Theorem 1 is true even if
is an obtuse, nonisosceles triangle. The
proof is adapted analogously. Theorem 2 (The Dual of the Theorem 1). If point in his plan, and triangle, as well as
is a triangle,
Bobillier transversals relative to a certain transversal in
perpendiculars in , and on
a certain of , and the
respectively, intersect the Bobillier
2
transversals in the points
, then the ceviens
are
Proof. We convert by duality with respect to a circle
the figure
concurrent.
indicated by the statement of this theorem, i.e. Figure 2. Let polars of the points
be the
with respect to the circle
To the lines
will correspond their poles To the points respectively
will
correspond
polars
their
concurrent in
transversal’s pole Since the
, it means that polars
and
are
perpendicular, so
, but
pass through that
Figure 2.
, which means
contains the height from of
triangle
similarly from
and
contains the height and
height from
contains the of
triangle. Consequently, the pole of
transversal is the orthocenter points
of
triangle. In the same way, to the
will correspond the polars to
respectively through
and are concurrent in a point
line
with respect to the circle
that the polara
and
also
are perpendicular,
Given
, the pole of the it means
correspond to the cevian
passes through the the pole of the transversal 3
which pass
, so through
, in other words , so
is perpendicular taken from is perpendicular taken from
on
; similarly,
on
To the cevian
will correspond by duality considered to its pole, which is the intersection of the polars of
and
namely the intersection of
with the perpendicular taken from
we denote this point by
, i.e. the intersection of lines
. Analogously, we get the points
and
and on
, ;
Thereby,
we got the configuration from Theorem 1 and Figure 1, written for triangle of orthocenter
Since from Theorem 1 we know that
collinear, we get the the cevians transversal
are
are concurrent in the pole of
with respect to the circle
and Theorem 2 is
proved.
References [1] Ion Patrascu, Florentin Smarandache: „The Dual Theorem concerning Aubert Line”. [2] Florentin Smarandache, Ion Patrascu: „The Geometry of Homological Triangles”. The Education Publisher Inc., Columbus, Ohio, USA – 2012.
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