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(Crying)

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- Jul 18th 2010, 10:44 PMearthboya triangle geometry...
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(Crying) - Jul 18th 2010, 11:28 PMsimplependulum
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 .