, 2006

Solution:

a. Let $M, N$ be the midpoints of $AB$ and $AC$ respectively; being the common chord of the two circles of diameters $AB$ and $AC$, $AE$ is perpendicular to $MN$. If $E\widehat{A}D = 90^\circ$, then $AD$ and $MN$, being perpendicular to the same line (the one containing $EA$), are parallel; finally, by Thales' theorem, $MN$ is parallel to $BC$ and hence $BC$ is parallel to $AD$.

b. Similarly to what was done in point (a), letting $L$ be the midpoint of $AD$, we have that $AF$ is perpendicular to $NL$ and, if $F\widehat{A}B = 90^\circ$, then $AB$ is also perpendicular to $AF$ and thus parallel to $NL$. Applying Thales' theorem as above, we obtain that $AB$ is parallel to $CD$. Therefore, if $E\widehat{A}D = F\widehat{A}B = 90^\circ$, we have that $AB$ is parallel to $CD$ and $AD$ is parallel to $BC$, that is, $ABCD$ is a parallelogram.

c. Conversely, if $ABCD$ is a parallelogram, letting $M, N, L$ be the midpoints of $AB, AC, AD$ as above, we have that $AD, BC, MN$ are parallel and hence all perpendicular to $AE$, the common chord of the circles of diameters $AB$ and $AC$ (and centers $M$ and $N$); in the same way, $AB, CD, NL$ are parallel and hence all perpendicular to $AF$, the common chord of the circles of diameters $AC$ and $AD$ (and centers $N, L$). Therefore $E\widehat{A}D = F\widehat{A}B = 90^\circ$.