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/sci/ - Science & Math

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8622555 No.8622555 [Reply] [Original]

Apollonius prop. 38

Trying to prove rect. FG,GH = sq. GC

Since
rect. FG, GH = square GC = rect. CG, GD
for
CG=GD,
therefore
rect. FG,GH = rect. CG,GD;
therefore
FG:GD :: CG:GH
and convertendo
GF:FD :: GC:CH
and let the doubles of the antecedents be taken; but
2GF= CF+FD
because
CG=GD

I shouldn't have to draw it out, G is the midpoint of both the regular diameter and the second diameter(CD) of the hyperbola. Everything else is just playing with ratios.

Now my question is how the FUCK isn't CF added to FD the same as CD???

>> No.8622566

The only other detail you'll need is that H is the end of the transverse of the conjugate diameter parallel to the diameter.