 # In geometry class, this property was quickly introduced: a/b=c/d=>(a+c)/(b+d)=a/b=c/d It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove. How to demonstrate it? Leila Jennings 2022-07-14 Answered
Why $\frac{a}{b}=\frac{c}{d}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{a+c}{b+d}=\frac{a}{b}=\frac{c}{d}$?
In geometry class, this property was quickly introduced:
$\frac{a}{b}=\frac{c}{d}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{a+c}{b+d}=\frac{a}{b}=\frac{c}{d}$
It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove.
How to demonstrate it?
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$\frac{a}{b}=\frac{c}{d}⇒\frac{a+c}{b+d}=\frac{a+\frac{ad}{b}}{b+d}=\frac{ab+ad}{b\left(b+d\right)}=\frac{a}{b}=\frac{c}{d}$

We have step-by-step solutions for your answer! Caylee Villegas
Assume that none of the unknowns is zero. Then, let
$\frac{a}{b}=\frac{c}{d}=k$
This implies,
$a=kb,\phantom{\rule{2em}{0ex}}c=kd$
$\therefore a+c=k\left(b+d\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{a+c}{b+d}=k$

We have step-by-step solutions for your answer!