# Triangle area equals quadrilateral area

Here is a diagram and challenge description that should be clear and simple to understand. • How did you find the midpoint without a compass? – KSmarts Feb 23 '15 at 18:06
• The midpoints are provided. – Moti Feb 23 '15 at 20:06
• I know that, I'm asking how you got them, and why can't we have the same tools? – KSmarts Feb 23 '15 at 20:07
• 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? – fibonatic Feb 23 '15 at 20:35
• It's easy if you use a compass - just use shear transformations. – Joe Z. Feb 23 '15 at 22:08

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.

• How do triangles DFB and DFX have the same height? – Marmy1954 Feb 25 '15 at 2:48
• Height of both is the distance between the parallel lines. – Joel Rondeau Feb 25 '15 at 2:53
• And how do you construct BX to be parallel with only an unmarked ruler? – Marmy1954 Feb 25 '15 at 3:07
• 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. – Joel Rondeau Feb 25 '15 at 3:10

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!

• The idea is of a ruler without marking. No measurement is possible, even not adding markings:) – Moti Feb 18 '15 at 5:20
• @Moti Hmmmm. back to the drawing board. – not my job Feb 18 '15 at 9:03