What is "the center of the arc $BC$ of the circumcircle of $\triangle ABC$ with point $A$"? Who is point $E$ ? What does it mean that "the projection of a point, on the line $AC$, bisects the length of the lines $BA$ and $AC$" ?

I didn't know how to translate it well, so I just translated it as written. By the center of the arc, I meant a point on a circle for which $BD=CD$, but because there are two points defined that way, we consider the one closer to $A$. For the other thing, what was meant by the author was to prove that $|CE|=\frac{|AC|+|AB|}{2}$. Hope you understood. Please edit my post, so others can understand too. Thank you.

i don't think you explained it well but i do know $E$ is the projection of $D$ on $AC$. now let $P$ be on the circumcircle of $\triangle ABC$ such that $DP$ is the diameter of that circle and let $F$ be the projection of $P$ on line $AC$. than first we have to prove $AF=(AB+AC)/2$ so let the circle with center $P$ and radius $PC$ intersect $AC$ once again at $Q$ it is is to prove that $\triangle APB \cong \triangle AQP$(using that $P$ is the center of arc $BC$ on the circumcircle of $\triangle ABC$ without $ A$) so $AB=AQ$ but $\triangle PCQ$ is isosceles so $FQ=FC$ now $AF=(AB+AC)/2$ now we have that $ADCP$ is cyclic with $DP$ as it's diameter and the projections of $D$ and $P$ on diagonal $AC$ are $E$ and $F$. we have to prove that $CE=AF$ which is a known problem(or at least for me). since $ADCP$ is cyclic with diameter $DP$ it's easy to prove $\triangle AFP \sim \triangle DCP$ so $AP/DP=AF/DC$ but also easy to prove is $\triangle ADP \sim \triangle DEC$ so $AP/DP=EC/DC$ and from the last 2 we have $AF=CE$ and done

It is just the broken chord theorem in fact. using wallace line and menelaos on perp. from D it is solved directly.

My solution in the competition went like this: Draw a circle $\omega$ through $D$ with radii $DC=DB$. Let $S$ be the second point of intersection of $\omega$ and the line $AC$. Because then $CE=\frac{CS}{2}$ it is enough to prove that $AS=AB$. But $\angle CAB=\angle CDB=2 \angle CSB\implies \angle ASB=\angle CSB=\angle ABS$ thus $AS=AB$. $\blacksquare$

Drop perpendiculars $DF$ and $DG$ respectively on $BC$ and $AB$ (perpendicular meets it at its extension). $E,F,G$ are by Simson's theorem collinear. By Menelaus' theorem $\frac{AG}{GB}\frac{BF}{FC}\frac{CE}{EA}=1$. By the problem condition $BF=FC$ and by some simple angle chase we get $AG=EA$. So $GB=CE$ which gives $AB+AE=CE$, or equivalently, $AB+AC=2CE$. Done

Easy problem. Let X be the point on AC such that XA=XB, <BXC=<a/2=<BDC/2 and DB=DC => D is the circumcenter of the triangle XBC, so DX=DC and DXC is iscoceles, so E is the midpoint of XC, so CE=XC/2=(b+c)/2.