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Let ABCD be a trapezium. AB is parallel CD. EF is a line parallel to AB and CD. E is on BC. F is on AD. What is the length of EF in termsof sides of trapezium? Given AC/CD=M/N=BF/FC. I derived the answer for special case in which the line joins the mid points of non parallel sides. It's mean of length of parallel sides in that case. I also know that length of line segment joining the mid points of diagonal of a trapezium is mean of difference of parallel sides. Does it help? Length of parallel sides is given.

Vindhyachal Vindhyachal
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  • Let the trapezium be $ABCD$ (counterclockwise) with $AB$ parallel and not smaller than $CD$. Draw $CE$ parallel to $AD$ with $E$ on $AB$. Assume the segment in question is called $FG$, with $F$ on $AD$ and $G$ on $CD$. Call the intersection of $FG$ and $CE$, $H$. We need the length of $FG$. This length is $FH+HG$. But $FH=DC$ (the top basis of the trapezium). Now the given ratio is $CG/CB=HG/EB=HG/(AB-DC)$. So $FG=DC+(CD/CB)(AB-DC)$. – Peyton Jul 30 '17 at 20:30

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Let $ABCD$ be our trapezoid, where $BC||AD$, $AD=a$, $BC=b$, where $a>b$.

$M\in AB$ and $N\in CD$ such that $MN||AD$, $MN=x$ and $AM:MB=n:m$.

Let $K\in AD$ such that $BK||CD$. Also, let $BK\cap MN=\{L\}$.

Thus, $KD=LN=BC=b$ and $$\frac{ML}{AK}=\frac{BM}{AB}$$ or $$\frac{x-b}{a-b}=\frac{m}{m+n}$$ or $$x=\frac{ma+nb}{m+n}.$$ Done!

Michael Rozenberg
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