Journal for Geometry and Graphics 29 (2025), No. 1, 043--053
Copyright by the authors licensed under CC BY SA 4.0
More Variations on Fermat Analogue of the Steiner-Lehmus Theorem
Sadi Abu-Saymeh
Concord, U.S.A., North Carolina, U.S.A.
ssaymeh@yahoo.com
Mowaffaq Hajja
Irbid, Jordan
mowhajja@yahoo.com
[Abstract-pdf]
The celebrated Steiner-Lehmus theorem states that if the internal bisectors of two angles of a triangle are equal, then the corresponding sides have equal lengths. In this paper, we consider the triangle ABC whose all angles are less than $120^{\circ}$, $F$ is its Fermat point, and $\text{per}(ABC)$, $[ABC]$ stand for its perimeter and area, respectively. In Theorem 1, we prove the Fermat analogue of Steiner-Lehmus Theorem that states that if the cevians from $B$ and $C$ through the Fermat point $F$ meet $AC$ and $AB$ at $B'$ and $C'$ respectively, then $BB' = CC'$ is equivalent to $AB = AC$. More stronger forms are also proved such as $AB > AC$ is equivalent to each of $BB' > CC'$ and $\text{per}(C'BC > \text{per}(B'CB)$. More variations on Fermat analogue of Steiner-Lehmus Theorem are proved in Theorems~3 and 4. In Theorem 3, the cevians through $F$ from $B$ and $C$ meet the external angle bisectors of $C$ and $B$ at $D$ and $E$ respectively, and it is proved that, for example, $AB = AC$ is equivalent to each of $CE = BD$, $\text{per}(EC'B) = \text{per}(DB'C)$, and $[EC'B] = [DB'C]$ and more stronger forms are also proved such as $AB > AC$ is equivalent to each of $CE > BD$, $\text{per}(EC'B) > \text{per}(DB'C)$, and $[EC'B] > [DB'C]$. In Theorem 4, we prove that if the angle $A$ of the triangle $ABC$ is not equal to $60^{\circ}$ and the circumcevians $BK$ and $CL$ of the Fermat point $F$, that meet the circumcircle of $\triangle ABC$ at $K$ and $L$, are equal, then the triangle $\triangle ABC$ is isosceles with $AB = AC$.
Keywords: Steiner-Lehmus Theorem, Fermat point, cevian, circumcevian.
MSC: 51M04.
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