# a triangle geometry...

• Jul 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.

• Jul 18th 2010, 11:28 PM
simplependulum
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 .