# a triangle geometry...

• July 18th 2010, 10:44 PM
earthboy
a 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.

• July 18th 2010, 11:28 PM
simplependulum
This is a problem of USSR MO , i have solved it already .

Let $M$ be the mid-pt of $BE$ , from $BD = CD$ we have $CE \parallel DM$ .

Then consider that $\Delta ADE \sim \Delta DBE$ so

$\Delta AFD \sim \Delta DMB$ because $F$ is the mid-pt of $DE$ while $M$ is the mid-pt of $BE$

We can find that $AF$ is perpendicular to $DM$ .

( Median perpendicular to median as sides perpendicular to sides , check it ! )

Therefore , $CE$ is perpendicular to $AF$ as $CE \parallel DM$ mentioned before .