Olimpiadi di Matematica - Febbraio

Solution:

Let $\alpha = \widehat{DAB} = \widehat{ABC}$ (since the trapezoid is isosceles, the two angles are equal); the condition $AB = AP$ is equivalent to having $\widehat{BPA} = \widehat{PBA} = \alpha$. In turn, since $\widehat{PAB} = \alpha / 2$ ($AP$ is the bisector of the angle at $A$), this is equivalent to $2\alpha + \alpha / 2 = 180^\circ$ (sum of the interior angles in triangle $BAP$), that is, to $\alpha = 72^\circ$.

Let us set $\widehat{CDB} = \widehat{BDA} = \theta$; since the angles $\widehat{CDB}$ and $\widehat{DBA}$ are alternate interior angles with respect to the parallels $DC, AB$ cut by the transversal $BD$, we also have $\widehat{DBA} = \theta$; hence, since the trapezoid is isosceles, we in fact also have $\widehat{BCA} = \widehat{CAB} = \theta$, and moreover $2\theta + \alpha = 180^\circ$ (interior angles of triangle $ABC$), that is $\theta = 90^\circ - \alpha / 2$.

Consequently, $\widehat{CAP} = \widehat{CAB} - \widehat{PAB} = \theta - \alpha / 2 = 90^\circ - \alpha$, while $\widehat{DAC} = \alpha - \widehat{CAB} = \alpha - \theta = \frac{3}{2}\alpha - 90^\circ$.

Now, $AC$ is the bisector of $\widehat{DAP}$ if and only if $\widehat{DAC} = \widehat{CAP}$, if and only if $\frac{3}{2}\alpha - 90^\circ = 90^\circ - \alpha$, that is $\alpha = 72^\circ$, which in turn is equivalent to the condition $AB = AP$ as argued previously.