Italian Mathematical Olympiad - February Round

Solution:

a.
Since $DBAE$ and $CDFA$ are cyclic, we have $\angle BCP = \angle DCF = \angle DAF = \angle DAB$ and $\angle CBP = \angle DBE = \angle DAE = \angle DAC$. Since $AD$ is the bisector of $\angle BAC$, we have $\angle BAD = \angle DAC$, hence $\angle BCP = \angle CBP$, which shows that triangle $BCP$ is isosceles.

b.
Since $DBAE$ and $CDFA$ are cyclic, we have $\angle DFC = \angle DAC = \angle DAB = \angle DEB$. Moreover, from the previous point, we know that $\angle EBD = \angle DCF$. Since triangles $EBD$ and $FCD$ have 2 pairs of equal angles, they are similar. Therefore $BD : BE = CD : CF$.

c.
The quadrilateral $CFBF^{\prime}$ has, by hypothesis, diagonals that bisect each other, hence it is a parallelogram. Note that $\angle EBF^{\prime}$ is congruent to $\angle CAB$. Indeed, since $CFBF^{\prime}$ is a parallelogram, we know that $\angle CBF^{\prime} = \angle BCF = \angle DAF$ (where the last equality follows from the previous points) and $\angle CBE = \angle DAE$ (again by the previous points). Hence
$$\angle EBF^{\prime} = \angle CBE + \angle CBF^{\prime} = \angle DAE + \angle DAB = \angle CAB .$$
Now observe that
$$\angle EC F^{\prime} = \angle ECB + \angle BCF^{\prime} = \angle ACB + \angle CBA = 180^{\circ} - CAB = 180^{\circ} - EBF^{\prime} \, ,$$
where the equality $\angle BCF^{\prime} = \angle CBA$ is due to the fact that the quadrilateral $CFBF^{\prime}$ is a parallelogram. It follows that the angles $\angle EC F^{\prime}$ and $\angle EBF^{\prime}$ are supplementary and the quadrilateral $EBF^{\prime}C$ is cyclic. We deduce therefore that $\angle ECB = \angle EF^{\prime}B$. We have thus shown that triangles $EF^{\prime}B$ and $ACB$ have two pairs of equal angles and are therefore similar.

Alternatively, we can observe that $BF^{\prime} = CF$ because they are opposite sides of the parallelogram $BF^{\prime}CF$, from which, thanks to point (b), we obtain $BF^{\prime} : BE = CD : BD$. Now, thanks to the angle bisector theorem, $CD : BD = AB : AC$, hence triangles $BEF^{\prime}$ and $ABC$ are similar thanks to the second criterion of similarity.