Let the parallel cuts $BC$ at $K$ and the angle bisector cuts $AC$ at $N$. We have to prove that $KC$=$\frac{a+c}{2}$. \[KC=\frac{a+c}{2} \Leftrightarrow\] \[MC \frac{\sin \angle KMC}{\sin \angle MKC} = \frac{a+c}{2}\Leftrightarrow\] \[\frac{\sin \angle HNC}{\sin (180^{\circ}- C-\angle HNC)} = \frac{a+c}{b} \Leftrightarrow\] \[\frac{\sin(90^{\circ}+\frac{ A - C}{2})}{\sin (90^{\circ}+\frac{A + C}{2})}=\frac{\sin A+\sin C}{\sin B}\Leftrightarrow\] \[\frac{\cos\frac{A-C}{2}}{\sin\frac{B}{2}}=\frac{\sin A + \sin C}{2\sin\frac{B}{2}\cos\frac{B}{2}} \Leftrightarrow\] \[2\sin\frac{A+C}{2}\cos\frac{A-C}{2} = \sin A + \sin C.\] But it is the compound angle formula.

I suppose AB>BC (if AB=BC it is evidently). I note $U\in AB\cap l$ and the point S where the bisector of the angle $\widehat{ABC}$ meets the line AC. It can be demonstrated easily that MU || BS (with equalities between angles) . From $AUM\sim ABS$ results: $\frac{c}{AU}=\frac{AS}{AM}=\frac{bc}{a+c}\cdot \frac{2}{b}=\frac{2c}{a+c}\Longrightarrow AU=\frac{a+c}{2},\ BU=\frac{c-a}{2}$. Thus$,\ CB+BU=a+\frac{c-a}{2}=\frac{a+c}{2}=AU,\ i.e.\ MC+CB+BU=MA+AU$.

Jutaro wrote: Let $ABC$ be a triangle, $H$ the orthocenter and $M$ the midpoint of $AC$. Let $\ell$ be the parallel through $M$ to the bisector of $\angle AHC$. Prove that $\ell$ divides the triangle in two parts of equal perimeters. Pedro Marrone, Panamá Proof. Construct the parallelogram $AUCD.$ Assume that: $AB>AC,$ and denote: $S$ be the foot of angle bisector of angle $\widehat{ABC},\, U=AB\cap l,\,V=BC\cap l$ Thus, easy show that: $BS\parallel MV\implies\widehat{BVU}=\widehat{CBS}=\widehat{SBA}=\widehat{MUA}=\widehat{BUV}=\widehat{MDC}$ $\implies\begin{cases}BU=BV\\ CV=CD\end{cases}\implies CB+BU=AU$ $\implies MC+CB+BU=MA+AU$

I will use the attached diagram; First note that $MK$ is common in both parts, similarly with $MC$ that is equal to $MA$, so, the real problem is to show that $BK+AB=KC$, some angle chasing with the parallel will show that $\angle BKY=\angle BYK$, meaning that $BK=BY$, now for the conclusion, we draw the parallel to $MY$ through $C$, then it is clear that $KC=YD$ and that $Y$ is the midpoint of $AD$, so $AB+BY=YD$ but with substitutions we have that $AB+BY=AB+BK$ so we are done

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