Proving Proportionality of Areas with Affine Geometry

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Homework Statement



Show that the ratio of areas is proportional to the sides squared:

[tex]\frac{[ACD]}_{[CDB]}[/tex] is proportional to [tex]\frac{AC^2}_{CB^2}[/tex]

Please, see the picture: http://dl.getdropbox.com/u/175564/geo_henry.JPG .

Homework Equations



AC = 2 * CB

[tex]\frac{AD}_{DB}[/tex] is propotional to [tex]\frac{[ACD]}_{[CDB]}[/tex]

where [ACD] and [CDB] are areas.

The Attempt at a Solution



I was unable to prove the relation with pythagoras, so I feel an easier solution. Perhaps, you can prove it somehow with affine geometry.
 
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Please, see the picture here: http://dl.getdropbox.com/u/175564/geo_henry.JPG . Or download the attachment.
 

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The basic idea is that if two triangles are similar (same corresponding angles), their corresponding sides will be proportional.

[itex]\angle[/itex]ACD = [itex]\angle[/itex]CDM - alt. interior angles cut by transversal
[itex]\angle[/itex]MDB = [itex]\angle[/itex]DAC - complements of congruent angles are congruent
[itex]\angle[/itex]ABC = [itex]\angle[/itex]ACD - complements of congruent angles are congruent

The statements above show that triangle ACD is similar to triangle CDB.
[ACD] = 1/2 * AD * CD
[CDB] = 1/2 * DB * CD

The two equations above show the proportionality you want.
 
Mark44 said:
The statements above show that triangle ACD is similar to triangle CDB.
[ACD] = 1/2 * AD * CD
[CDB] = 1/2 * DB * CD

The two equations above show the proportionality you want.

I am sorry of the blurry image. I wanted to know why the relation is true:

[tex]\frac{AC^2}_{CB^2}[/tex] is proportional to [tex]\frac{[ACD]}_{[CDB]}[/tex]
 

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