Estonia — Open Contests

Let $D$ be the point of intersection of $BC$ and median from vertex $A$, $E$ be the point of intersection of angle bisector from vertex $B$ and $AC$, and $F$ be the point of intersection of altitude from vertex $C$ and $AB$ (see fig. 6). As $\angle FBC = 45^\circ$ and $\angle CFB = 90^\circ$, triangle $FBC$ is a right isosceles triangle with $|CF| = |FB|$.

Figure 6

Let $X$ and $Y$ be the points of intersection of the line that passes through point $D$ and is parallel to $CF$, with lines $AC$ and $AF$, respectively. We have $\angle YAX = 22.5^\circ$, $\angle XYA = 90^\circ$ and $\angle AXY = 180^\circ - \angle YAX - \angle XYA = 67.5^\circ$. Therefore
$$\angle XCD = 180^\circ - \angle BCA = 67.5^\circ = \angle AXY = \angle CXD,$$
due to which $|XD| = |CD| = |DB|$.
As line segments $DY$ and $CF$ are parallel, $DY$ is the midsegment of triangle $BCF$. Hence
$$\frac{|XD|}{|DY|} = \frac{|DB|}{|DY|} = \frac{|CB|}{|CF|} = \frac{|CB|}{|FB|}.$$
Let now $K$ be the point of intersection of $BE$ and $CF$. The angle bisector property gives that $\frac{|CK|}{|KF|} = \frac{|CB|}{|FB|}$, so $\frac{|XD|}{|DY|} = \frac{|CK|}{|KF|}$, from which $\angle FAK = \angle YAD$. Therefore also $AD$ passes through $K$, QED.

Figure 7

Similarly to the previous solution we pick triangle $ABC$, mark points $D$, $E$ and $F$ and show that $|CF| = |FB|$. Additionally notice that $\angle ABE = \frac{\angle ABC}{2} = \frac{45^\circ}{2} = 22.5^\circ = \angle BAE$, which gives $|BE| = |AE|$.
Let now $Z$ be a point on $BC$ such that $\angle BEZ = 90^\circ$ (see fig. 7). Then $\angle ZBE = \frac{\angle ABC}{2} = 22.5^\circ = \angle CAF$ and $\angle BEZ = 90^\circ = \angle AFC$, hence triangles $BEZ$ and $AFC$ are
$$\angle EZC = 180^\circ - \angle ZBE - \angle BEZ = 67.5^\circ = 180^\circ - \angle BCA = \angle ECZ,$$
which gives $|EZ| = |EC|$. Thus $\frac{|AE|}{|EC|} = \frac{|BE|}{|EZ|} = \frac{|AF|}{|FC|} = \frac{|AF|}{|FB|}$. Therefore $\frac{|AF|}{|FB|} \cdot \frac{|BD|}{|DC|} = \frac{|CE|}{|EA|} = 1$ and Ceva's theorem gives that $AD$, $BE$ and $CF$ intersect in one point.

Similarly to first solution we choose triangle $ABC$, mark points $D$, $E$, $F$ and $Y$ and show that $|CF| = |FB|$.
Let the points of intersection of $CF$ with angle bisector $BE$ and median $AD$ be $K_1$ and $K_2$, respectively. Let $h_1 = |FK_1|$ and $h_2 = |FK_2|$ and in addition let $u = |AF|$ and $v = |CF| = |FB|$. Let us show that $h_1 = h_2$, then $K_1 = K_2$. Using the similarity of triangles $BFK_1$ and $CFA$, similarity of triangles $BCK_1$ and $ABC$ and similarity of triangles $AFK_2$ and $AYD$, we get the equalities
$$\frac{h_1}{v} = \frac{v}{u}, \quad \frac{v-h_1}{\sqrt{2}v} = \frac{\sqrt{2}v}{u+v}, \quad \frac{h_2}{u} = \frac{\frac{1}{2}v}{u+\frac{1}{2}v}$$
which imply $\frac{v^2}{u} = h_1 = \frac{(u-v)v}{u+v}$ and $h_2 = \frac{uv}{2u+v}$. From equation $\frac{v^2}{u} = \frac{(u-v)v}{u+v}$ we get $2u + v = \frac{u^2}{v}$, therefore $h_2 = \frac{uv}{2u+v} = \frac{v^2}{u} = h_1$.