Italy competition problems

Solution:

For convenience let us call $\widehat{BCM} = \alpha$. Let us consider triangles $\triangle ADC$ and $\triangle BAC$: they have two congruent angles, and hence also the third is congruent by difference ($\widehat{CAD} = \widehat{CBA}$). By the first similarity criterion they are therefore similar; in particular we have $AD : BA = CA : CB$. Letting $N$ be the midpoint of $AD$, let us consider triangles $\triangle CAN$ and $\triangle CBM$, which are similar since they have $AN : BM = 2AN : 2BM = AD : BA = CA : CB$ and $\widehat{CAN} = \widehat{CBM}$ as shown before. In particular we have that $\widehat{ACN} = \widehat{BCM} = \alpha$ as corresponding angles of similar triangles

from which
$$\widehat{NCM} = \widehat{ACN} + \widehat{ACM} = \widehat{BCM} + \widehat{ACM} = \widehat{ACB}.$$
Let us consider the quadrilateral $ANCM$ and prove that it is cyclic (inscribable in a circle): summing the opposite angles at $A$ and $C$ we obtain
$$\widehat{NAM} + \widehat{NCM} = (\widehat{DAC} + \widehat{CAB}) + \widehat{ACB} = \widehat{CBA} + \widehat{CAB} + \widehat{ACB} = 180^\circ$$
where the last equality follows from the fact that the sum of the interior angles of a triangle is a straight angle. From the cyclicity of $ANCM$ we deduce $\widehat{NMA} = \widehat{NCA} = \alpha$ (angles subtending the same arc). But now, since $N$ and $M$ are the midpoints of $AD$ and $AB$, by Thales' theorem we have that $NM \parallel BD$; in conclusion $\widehat{DBA} = \widehat{NMA} = \alpha$, since they are corresponding angles of two parallel lines cut by a transversal.

Solution:

Let us consider $E$ such that $EACD$ is a parallelogram and let us prove that triangles $EAC$ and $DAB$ are similar (using the results from the first part of the solution shown above).

Solution:

Let us consider the Gauss plane where $c = 0$ is the origin and $b = 1$. $\triangle ADC \sim \triangle BAC$ algebraically translates into $d = a^2$.

Expressing the claim in terms of complex numbers we must show that $\arg \frac{(a+b)/2 - c}{b-c} = \arg \frac{b-d}{b-a}$; carrying out the computations we obtain:
$$\arg \frac{(a+b)/2 - c}{b-c} = \arg \frac{z+1}{2} = \arg(z+1) = \arg \frac{1-z^2}{1-z} = \arg \frac{b-d}{b-a}$$