Prof. Ion Pătraşcu – The National College “Fraţii Buzeşti”, Craiova, Romania Prof. Florentin Smarandache – University of New Mexico, U.S.A.

Abstract. In this article we’ll emphasize on two triangles and provide a vectorial proof of the fact that these triangles have the same orthocenter. This proof will, further allow us to develop a vectorial proof of the Stevanovic’s theorem relative to the orthocenter of the Fuhrmann’s triangle.

Lemma 1

Let ABC an acute angle triangle, H its orthocenter, and ‘, ‘, ' A B C the symmetrical points of H in rapport to the sides , , BC CA AB .

We denote by , , X Y Z the symmetrical points of , , A B C in rapport to ' ‘, ' ‘, ' ' B C C A A B The orthocenter of the triangle XYZ is H .

      A 

C’

        V      

X

B’

H Y

Z

   U  

      C 


 B 













        A’ 

Fig. 1

Proof

We will prove that XH YZ ⊥ , by showing that 0 XH YZ ⋅

JJJG JJG . We have (see Fig.1)

VH AH AX

− JJJG JJJG JJJG

BC BY YZ ZC

JJJG JJJG JJG JJG ,
from here

YZ BC BY ZC

− − JJG JJJG JJJG JJG

Because Y is the symmetric of B in rapport to ' ' A C and Z is the symmetric of C in rapport to ' ' A B , the parallelogram’s rule gives us that:

' ' BY BC BA

JJJG JJJG JJJG

' ' CZ CB CA

JJG JJJG JJJG .

Therefore

( ) ' ' ' ' YZ BC BC BA B C A C

− + + + JJG JJJG JJJG JJJG JJJJG JJJJG

But

' ' BC BH HC

JJJG JJJG JJJJG

' ' BA BH HA

JJJG JJJG JJJG

' ' CB CH HB

JJJG JJJG JJJG

' ' CA CH HA

JJJG JJJG JJJG

By substituting these relations in the YZ JJG , we find:

' ' YZ BC C B

JJG JJJG JJJJG

We compute
( ) ( ) ' ' ' ' ' ' XH YZ AH AX BC C B AX BC AH C B AX BC AX C B ⋅

− ⋅ +

⋅ + ⋅ − ⋅ − ⋅ JJJG JJG JJJG JJJG JJJG JJJJG JJJG JJJG JJJG JJJJG JJJG JJJG JJJG JJJJG

Because AH BC ⊥

we have

0 AH BC ⋅

JJJG JJJG ,
also
' ' AX B C ⊥

and therefore
' ' 0 AX B C ⋅

JJJG JJJJG .

We need to prove also that

' ' XH YZ AH C B AX BC ⋅

⋅ − ⋅ JJJG JJG JJJG JJJJG JJJG JJJG

We note:

{ } U AX BC

∩ and { } ' ' V AH B C

∩

( ) ( ) , AX BC AX BC cox AX BC AX BC cox AUC ⋅

⋅ ⋅

⋅ ⋅ JJJG JJJG ) )

( ) ( ) ' ' ' ' , ' ' ' ' ' AH C B AH C B cox AH C B AH C A cox AVC ⋅

⋅ ⋅

⋅ ⋅ JJJG JJJJG ) )

We observe that
' AUC AVC ≡ ) ) (angles with the sides respectively perpendicular).

The point ' B is the symmetric of H in rapport to AC , consequently
' HAC CAB ≡ ) ) ,
also the point ' C is the symmetric of the point H in rapport to AB , and therefore

' HAB BAC ≡ ) ) .
From these last two relations we find that
' ' 2 B AC A

) ) .

The sinus theorem applied in the triangles ' ' AB C and ABC gives:

' ' 2 sin 2 2 sin B C R A BC R A

⋅

We’ll show that
' ' AX BC AH C B ⋅

⋅ ,
and from here
2 sin 2 sin 2 AX R A AH R A ⋅

⋅ ⋅

which is equivalent to
2 cos AX AH A

We noticed that
' ' 2 B AC A

) ,
Because

' ' AX B C ⊥ ,
it results that
TAB A ≡ ) ) ,
we noted { } ' ' T AX B C

∩ .
On the other side 1 ' ,
2 AC AH AT AY

= ,
and
‘cos cos AT AC A AH A

= ,
therefore

0 XH YZ ⋅

JJJG JJG .

Similarly, we prove that
YH XZ ⊥ ,
and therefore H is the orthocenter of triangle XYZ .

Lemma 2

Let ABC a triangle inscribed in a circle, I the intersection of its bisector lines, and
‘, ‘, ' A B C the intersections of the circumscribed circle with the bisectors , , AI BI CI respectively.

The orthocenter of the triangle ' ' ' A B C is I .

      A 






    C’ 

B’

I

B

      C 


  













        A’ 

Fig. 2

Proof

We’ll prove that ' ' ' A I B C ⊥ .

Let
q ( ) q ( ) ' ' m A C m A B α =

,
q ( ) q ( ) q ( ) q ( ) ' ' ' ' m B C m B A m C A m C B β γ

=

=

Then

( ) ( ) 1 ' ' 2 m A IC α β γ

)

Because
( ) 2 360 α β γ + +

°
it results
( ) ' ' 90 m A IC

° ) ,
therefore
' ' ' A I B C ⊥ .

Similarly, we prove that
' ' ' B I A C ⊥ ,
and consequently the orthocenter of the triangle ' ' ' A B C is I , the center of the circumscribed circle of the triangle ABC .

Definition

Let ABC a triangle inscribed in a circle with the center in O and ‘, ‘, ' A B C the middle of the arcs p p p , , BC CA AB respectively. The triangle XYZ formed by the symmetric of the points ‘, ‘, ' A B C respectively in rapport to , , BC CA AB is called the Fuhrmann triangle of the triangle ABC .

Note

In 2002 the mathematician Milorad S