How do I deduce the ratio of $AB:BC:CD$?.

Question

The line segment is given and points $A,B,C,D$ are marked on the line segment(endpoints of it are A,D respectively and $ B < C $) .

So there are $3$ segments at the line segment and We know the $2$ following ratios.

$AC:CD=1:s$

$AB:BD=t:1$

$s,t > 0$

From above ratios,the $AB:BC:CD$ is equal to $\displaystyle\frac{t}{1+t}:1-\displaystyle\frac{s}{1+s}-\displaystyle\frac{t}{1+t}:\frac{s}{1+s}$ $\quad $.

How do I deduce it?.

Suppose the length AD is L. Then $AB=\frac{t}{t+1}L$ and $CD=\frac{s}{s+1}L$. Hence $BC=L-AB-CD$. — almagest, Jan 25 '20 at 10:52
@almagest: Since your comment answers the OP, I suggest you write that out as an actual response so that the question does not remain unanswered. — Jose Arnaldo Bebita Dris, Jan 25 '20 at 11:33

score 1 · Accepted Answer · answered Jan 25 '20 at 11:38

1

Suppose the length $AD$ is $L$. Then $AB=\frac{t}{1+t}L$ and $CD=\frac{s}{1+s}L$, so $BC=L-AB-CD=L-\frac{t}{1+t}L-\frac{s}{1+s}L$. Hence the ratio is $$AB:BC:CD=\frac{t}{1+t}:1-\frac{t}{1+t}-\frac{s}{1+s}:\frac{s}{1+s}$$

answered Jan 25 '20 at 11:38

almagest

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How do I deduce the ratio of $AB:BC:CD$?.

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