let in triangle ABC,AB=AC. D is the midpoint of BC.E is the foot of the perpendicular from D to AB,and F is the midpoint of DE.
Prove that AF is perpendicular to CE.
please help
This is a problem of USSR MO , i have solved it already .
Let $\displaystyle M $ be the mid-pt of $\displaystyle BE $ , from $\displaystyle BD = CD $ we have $\displaystyle CE \parallel DM $ .
Then consider that $\displaystyle \Delta ADE \sim \Delta DBE $ so
$\displaystyle \Delta AFD \sim \Delta DMB $ because $\displaystyle F $ is the mid-pt of $\displaystyle DE $ while $\displaystyle M $ is the mid-pt of $\displaystyle BE$
We can find that $\displaystyle AF $ is perpendicular to $\displaystyle DM $ .
( Median perpendicular to median as sides perpendicular to sides , check it ! )
Therefore , $\displaystyle CE $ is perpendicular to $\displaystyle AF $ as $\displaystyle CE \parallel DM $ mentioned before .