Difficult geometry question Watch

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john !!
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Point E is selected on side AB of a traingle ABC in such a way that AE : EB = 1 : 3, and point D is selected on side BC in such a way that CD : DB = 1 : 2. The point of intersection of AD and CE is F. Find EF/FC + AF/FD.
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El Stevo
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irrelevant, but did you make this up or not?
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john !!
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Nope, textbook question.
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Gauss
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Are you asking for the sum of the ratios?
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john !!
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Yeah.
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El Stevo
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Draw a line parallel to AB. Call it DGH where D is point D, G is the interesect of the line EC and H is where the line hits the line AC.

Using similar triangles:

DG:3a = b:3b so...

DG = a
a = EA

EF = FG
AF = FD
(AF/FD) = 1.

DH:4a = b:3b so...

DH = 4a/3
GH = DH - DG = a/3

GC = (1/3)*EC
EG = (2/3)*EC

EF = FG
FC = (2/3)*EC

(EF/FC) = (1/2)

(EF/FC) + (AF/FD) = (1/2) + 1 = (3/2).
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john !!
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Yep, that's correct
and pretty amazing...
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Gauss
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(Original post by mik1a)
Yeah.
Take the case for when the triangle is at right angles.

By trig, you get EC = {y= 3-x} and AD = {y= 4-2x}

=> They intersect at the point (1,2).

Now using Pythagoras we deduce that |EF| = rt2 and |FC| = rt8

Similarly, |AF| = rt5 and |FD| = rt5.

The sums of the two ratios are thus

rt2/rt8 + rt5/rt5
=> 1/2 + 1
=> 3/2.

Due to the triangle law, these properties will work for any shaped triangle, hence you can be assured the ratio is as such for any triangle with the properties stated above.

Euclid
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