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February 19th, 2013, 06:13 AM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Area of triangle AEF
[attachment=0:2yqq22o7]area of triangle AEF.jpg[/attachment:2yqq22o7]

February 19th, 2013, 01:38 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,627 Thanks: 2077 
S = BC·CD  a  b  c = (BE + EC)(CF + FD)  BE(CF + FD)/2  EC·CF/2  (BE + EC)FD/2 = BE·CF/2 + EC·CF/2 + EC·FD/2 S² = (BE²CF² + CF²EC² + FD²EC²)/4 + (BE·CF²EC + BE·CF·FD·EC + CF·FD·EC²)/2 (a + b + c)²  4ac = (BE(CF + FD)/2 + EC·CF/2 + (BE + EC)FD/2)²  BE·FD(CF + FD)(BE + EC) = (BE(CF + 2FD)/2 + EC·CF/2 + EC·FD/2)²  BE·FD(CF + FD)(BE + EC) = (BE²CF² + CF²EC² + FD²EC²)/4 + (BE·CF²EC + BE·CF·FD·EC + CF·FD·EC²)/2 The required result follows. 
February 19th, 2013, 03:44 PM  #3 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Area of triangle AEF
I'm consistently amazed and impressed by these problems by Albert Teng and the answers from those who explain them. Have you considered NOT posting the answers right away, Albert? 
February 19th, 2013, 03:50 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,627 Thanks: 2077 
I was doing this kind of answer when I was aged twelve. I wonder whether Albert.Teng has a neater method.

February 19th, 2013, 04:34 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,413 Thanks: 1024  Re: Area of triangle AEF
Let k = length(AB) * length(AD). S = k  a  b  c 
February 19th, 2013, 10:52 PM  #6  
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: Area of triangle AEF Quote:
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