37. Own extension of the Steiner-Lehmus' problem.

by Virgil Nicula, May 8, 2010, 6:49 PM

Virgil Nicula wrote:
Extension of Steiner-Lehmus' problem. Let $\triangle ABC$ and let $D\in (BC)$ for which the ray $ [AD$ is the bisector of the angle

$\widehat{BAC}$. For $L\in (AD)$ denote $ M\in AC\cap BL$ and $ N\in AB\cap CL$. Prove that $BM=CN$ $\Longleftrightarrow$ $AB=AC$.

Proposed problem. Acute angled triangle $ABC$ is given. A line $l$ parallel to side $AB$ passing through vertex $C$ is drawn. Let the angle bisectors of $\angle{BAC} $

and $\angle{ABC}$ intersect the sides $BC$ and $AC$ at points $D$ and $F$ and line $l$ at points $E$ and $G$ respectively. Prove that if $DE=GF$ then $AC=BC$.


Proof.Define the following relation between any two real numbers $x$ , $y\ :\ x\ \mathrm{.s.s.}\ y\ \iff\ x=y=0\ \ \vee\ \ xy>0$ (same sign) .

Denote the length $l_a$ of the $A$-bisector a.s.o. Prove easily that $\boxed{(l_a-l_b)\ .s.s.\ (b-a)}\ (*)$ and $DE=FG\iff$ $bl_a=al_b\ (1)$ .

In conclusion, $b<a\stackrel{(*)}{\iff} l_a<l_b\stackrel{(1)}{\iff} b>a\ \mathrm{abs.}$ and $b>a\stackrel{(*)}{\iff} l_a>l_b\stackrel{(1)}{\iff} b<a\ \mathrm{ abs.}$

=====================================================================================

$(*)$ Indeed, $l_a=\frac {2\sqrt{bcs(s-a)}}{b+c}$ a.s.o. and $l_a-l_b\ \mathrm{.s.s.}\ (a+c)\sqrt{b(s-a)}-$ $(b+c)\sqrt{a(s-b)}\ \mathrm{.s.s.}\ b(s-a)(a+c)^2-$

$a(s-b)(b+c)^2=(b-a)\left[s(c^2-ab)+ab(a+b+2c)\right]\equiv (b-a)\cdot E$ . Observe that $(l_a-l_b)\ \mathrm{.s.s.}\ (b-a)$ because

$E\ \mathrm{.s.s.}\ (a+b+c)(c^2-ab)+2ab(a+b+2c)=c^3+c^2(a+b)+3abc+ab(a+b)>0$ .



Proposed problem. Let $ABC$ be a triangle and let $D\in (BC)$ be the intersection of $A$-symmedian with the sideline $BC$ .

For a point $L\in (AD)$ denote $E\in BL\cap AC$ , $F\in CL\cap AB$ . Prove that $BE=CF\ \iff\ AB=AC$ .

Remark. In particular case when $L:=S$ - the symmedian point of $\triangle ABC$ obtain a similar result with Steiner-Lehmus' theorem for the incenter $I$ .
This post has been edited 9 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:43 AM

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