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T-N-T (Posted on 2017-11-25) Difficulty: 3 of 5

  
Let ABC be an arbitrary triangle with points D, E, and F
lying on rays AB, BC, and CA respectively such that

   AD/AB = BE/BC = CF/CA = x.

Prove that Area(ΔDEF)/Area(ΔABC) = 3x2 - 3x + 1.
  

See The Solution Submitted by Bractals    
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Solution Solution | Comment 1 of 2
The three corner triangles have the same area because if you take any two of them their bases and heights are in inverse proportion.

Area(ΔAFD)=Area(ΔBDE)=Area(ΔCEF)
So
Area(ΔDEF)=Area(ΔABC)-3*Area(ΔAFD)
Let AC=b be the base of ΔABC and the height=h, area = bh/2 
The base and height of ΔAFD are then (1-x)b and xh, area = x(1-x)bh/2

Area(ΔDEF) = bh/2 - 3x(1-x)bh/2

The ratio sought is then 1 - 3x(1-x) = 3x2 - 3x + 1



  Posted by Jer on 2017-11-26 11:16:42
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