Here is a diagram and challenge description that should be clear and simple to understand.
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$\begingroup$ How did you find the midpoint without a compass? $\endgroup$– KSmartsFeb 23, 2015 at 18:06
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$\begingroup$ The midpoints are provided. $\endgroup$– MotiFeb 23, 2015 at 20:06
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$\begingroup$ I know that, I'm asking how you got them, and why can't we have the same tools? $\endgroup$– KSmartsFeb 23, 2015 at 20:07
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$\begingroup$ So the ruler can only be used to draw lines between two points and the intersection of any two lines can be viewed as a point? $\endgroup$– fibonaticFeb 23, 2015 at 20:35
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$\begingroup$ It's easy if you use a compass - just use shear transformations. $\endgroup$– user88Feb 23, 2015 at 22:08
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2 Answers
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Make a line segment starting at $A$, through $D$.
Make a line segment starting at $C$, through $B$.
Call the point where they intersect $G$.
The triangle $EFG$ has the same area as the quadrilateral.
Proof: Extend $ED$ and draw a line segment starting at $B$ and running parallel to $CD$, call the point where they intersect $X$.
I know that triangle $EFX$ has the same area as the quadrilateral because triangle $DFB$ and triangle $DFX$ have the same area (same base and height). Unfortunately, I can't draw line $BX$, but I can use the same proof by splitting the quadrilateral into 2 quadrilaterals.
Point $E$ bisects segment $AB$. Therefore, if I draw a line parallel to $AD$ starting at $E$, it will bisect $DB$.
Similarly, point $F$ bisects segment $CD$. Therefore, if I draw a line parallel to $CB$ starting at $F$, it will bisect $BD$.
Call that point $H$ and draw a line starting at point $G$, through $H$ until it intersects $EF$. Call that point $I$.
$DG$ is parallel to $EH$ and therefore $EIG$ has the same area as $IEDH$.
$BG$ is parallel to $FH$ and therefore $IFG$ has the same area as $IFBH$.
Together this means that $EFG$ has the same area as $EFBD$.
The lines drawn in the proof can't be drawn using just a ruler, but that is unimportant.
answered Feb 24, 2015 at 21:53
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$\begingroup$ How do triangles DFB and DFX have the same height? $\endgroup$ Feb 25, 2015 at 2:48
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$\begingroup$ Height of both is the distance between the parallel lines. $\endgroup$ Feb 25, 2015 at 2:53
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$\begingroup$ And how do you construct BX to be parallel with only an unmarked ruler? $\endgroup$ Feb 25, 2015 at 3:07
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$\begingroup$ You can't. You only need to get point G with the ruler. Everything after that is the proof. It doesn't need to be drawn. $\endgroup$ Feb 25, 2015 at 3:10
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If I am allowed to measure distances with the ruler, then it would work to
measure the distance from the crossing point if the two long lines to the point B and replicate that distance from E towards A, calling the new point G. Since FDG has the sum of the perpendicular heights of the two triangles that have FD as a base, it has the same area as the quadrilateral.
Of course if you mean ruler as in the classical ruler and compass construction, without the compass, I'm more stuck!
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$\begingroup$ The idea is of a ruler without marking. No measurement is possible, even not adding markings:) $\endgroup$– MotiFeb 18, 2015 at 5:20
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