Construction of a line making equal intercepts on $AC,AB$ on opposite sides of $BC$.

Question

In triangle $ABC$, $D$ is an arbitary point on side $BC$. I need to construct a line that pass through $D$ such that $CF=EB$. (point $F$ lies on the extension of of side $AC$)

By Menelaus's theorem

$$x=\frac{bc-da} {d} $$

we can construct the line from this information.

But my question is how can I construct the line without calculating segment $x$.

I think there is an easier way. Any ideas?

Answer 1

The Menelaus' relation can be rearranged as $$\frac{a+x}{b}=\frac{c}{d}$$

So the problem reduces to dividing $a+x+b$ in the ratio $c:d$. That is we have to divide two intersecting line segments in same ratio.

The following construction procedure naturally follows :

• Locate $G$ on $AC$ beyond $C$ such that $CG=AB$
• Draw a line through $G$ parallel to $CB$ and find $H$ so that $GH=CB$ (or complete the parallelogram $GCBH$).
• Also $D$ is known. Draw a line through $D$ parallel to $AG$. This meets $GH$ in $I$.
• Finally join $AH$ and draw a line through $I$ parallel to $AH$.
• This line meets $AG$ in $F$. Extend $DF$ to meet $AB$ in $E$.