We consider a problem from the Magazine in the picture, page 140.
"C:2488. In triangle $ABC\;(AB=AC)$ , let $M$ be the midpoint of side $(AC)$.
If $m(\angle MBC)=30^{\circ}$ , show that triangle $ABC$ is equilateral.
{Author : } I. TUDOR, Băbana, Argeș"
SOLUTION CiP
The figure in the problem is :
To this we have the properties :
$AM=MC=\frac{AC}{2}=\frac{AB}{2}\tag{1}$
$\widehat{MBC}=30^{\circ} \tag{2}$
The following Lemma is related to the geometric object Medians of a Triangle :
Lemma If a triangle has two equal medians then it is isosceles.
Proof of Lemma In the geometric object in the figure below (which is
subordinate to the more general object MIDDLE LINE) we have the properties :
$AM=MC=\frac{AC}{2}\;\;,\;\;BD=DC=\frac{BC}{2} \tag{$L_1$}$
$AG=\frac{2}{3}AD\;,\;DG=\frac{1}{3}AD\;\;;\;\;BG=\frac{2}{3}BM\;,\;MG=\frac{1}{3}BM \tag{$L_2$}$
From the hypothesis we have $AD=BM\;;\;(L_2)\Rightarrow AG=BG\;and\;MG=GD$. From here and from the equality of vertical angles $\widehat{AGM}=\widehat{BGD}$ results with the SAS case of congruence of triangles that
$\Delta AGM=\Delta BGD \tag{T}$
Then $(T)\Rightarrow AM=BD\;\underset{(L_1)}{\Rightarrow} AC=BC.$
qed Lemma $\square$
Let us consider the symmetry of the problem figure with respect to the line $BC$.
The point $A'$ is the symmetric point of $A$ , $M'$ of $M$. Then
$AA' \perp BC\;,\;MM'\perp BC\;\;,\;\;BM'=BM\;\;,\;\;\widehat{M'BC}=\widehat{MBC}$
In particular, the height $AD$ in the isosceles triangle $ABC$ is also the median.
First, triangle $MBM'$ is equilateral ($\widehat{MBM'}=60^{\circ}$), so
$MM'=BM \tag{3}$
Secondly, $ME$ being parallel to $AD$ (both perpendiculat to $BC$), and $M$ the midpoint of $AC$ , we have from the Midline Theorem that
$\frac{AD}{2}=ME=\frac{MM'}{2}\underset{(3)}{=}\frac{BM}{2}$
hence $AD=BM$.
Therefore, in triangle $ABC$ , medians $AD\; and\; BM$ are equal, so by the Lemma, $CA=CB.$ Triangle $ABC$ is therefore equilateral.
$\blacksquare$
