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Gabi's Proof of Li 📂Lemmas

Gabi's Proof of Li

Theorem

If bdf(b+d)0bdf(b+d)\neq 0 then ab=cd=ef    a+cb+d=ef \frac { a }{ b }=\frac { c }{ d }=\frac { e }{ f } \implies \frac { a+c }{ b+d }=\frac { e }{ f }

Description

“Gabi” is nothing else but a word made from two Hanja characters: add 加 and compare 比. Here, the compare 比 is the same as the ‘ratio’ in ratios, making it a theorem where everything is encapsulated in the name.

Proof

ab=cd=ef \frac { a }{ b }=\frac { c }{ d }=\frac { e }{ f }

Therefore, ab=ef\frac { a }{ b }=\frac { e }{ f } and cd=ef\frac { c }{ d }=\frac { e }{ f }. If we multiply both sides of ab=ef\frac { a }{ b }=\frac { e }{ f } by bfbf,

cd=ef \frac { c }{ d }=\frac { e }{ f }

and, if we multiply both sides of cd=ef\frac { c }{ d }=\frac { e }{ f } by dfdf,

cf=de cf=de

Adding up the two obtained equations on both sides gives us

(a+c)f=(b+d)e (a+c)f=(b+d)e

Dividing both sides by (b+d)f(b+d)f results in

a+cb+d=ef \frac { a+c }{ b+d }=\frac { e }{ f }