Squares, midpoints and heights Let $ABC$ be a traingle, we draw squares on the sides $AB$ and $AC$, now we draw a segment from the vertexes of the square which are closer and then it forms a triangle, so prove that the line throw A and the mid point of BC is perpendicular to the segment defined before.
I have a trigonometric proof, but I would like to know any synthetic solution. Thanks

Let the new segment be $DE$ and let $F$ be the point where the
perpendicular from $A$ intersects the line through $D$ and $E$.
By constructing triangles congruent to $\triangle ADF$ and 
$\triangle AEF$, one can show that $B$ and $C$ are both at distance
$AF$ from the line through $A$ and $F$
and on opposite sides of that line, 
so the midpoint of $BC$ is on that line. 
Since two points determine a line,
whatever line you construct through $A$ and the midpoint of $BC$ must
be that same line, which we already know is perpendicular to $DE$.
In this proof, I would handle the case where $F$ is outside the segment $DE$ separately from the case where $F$ is between $D$ and $E$, but maybe there's a way to combine the cases.
