Bottema with Similar Triangles

27 May 2014, Created with GeoGebra

Assuming the point names as complex numbers, the similarity of the three triangles could be expressed as

$\displaystyle\frac{T-E}{B-E}=\frac{T-A}{D-A}=\frac{C-F}{D-F}=u,$

for a complex number $u.$ From here,

$\displaystyle\frac{(T+C)-(E+F)}{(B+D)-(E+F)}=u,$

implying

$\displaystyle\frac{(A+C)-(E+F)}{(A+B)-(E+F)}=u,$

which exactly means that $E+F$ and, with it, $\displaystyle\frac{E+F}{2}$ is independent of $D.$

This problem is a follow-up of the one posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page. The simple solution above has been posted by Leo Giugiuc (Romania) who also observed that Dao's problem extends to the one above. Here's Dao's formulation:

Triangles $BET,$ $TAD,$ $DFC$ are similar isosceles. Prove that the midpoint $K$ of $EF$ is independent of the position of $D.$