perplexus.info :: Geometry : T-N-T

T-N-T (Posted on 2017-11-25)

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

| 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)

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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