Given the quadrilateral ABCD (see drawing) with mid diagonals, E of AC and F of BD, you need to create 5 polygons equal in area to 1/4 of the quadrilateral ABCD area by means of a straight-edge, drawing 9 lines. The 5 polygons - quadrilaterals or triangles - may partially overlap.

enter image description here

  • $\begingroup$ Are E and F both inside ABCD? Can we use the intersection of AC and BD? $\endgroup$
    – Dr Xorile
    Commented Dec 14, 2015 at 3:56
  • $\begingroup$ Yes and yes as the drawing shows. You may use any well defined point that results for intersection of two lines. $\endgroup$
    – Moti
    Commented Dec 14, 2015 at 4:11
  • $\begingroup$ Can the polygon be "pinched off", like two triangles with one vertex in common? Or like ACDB (note the order) in your diagram? $\endgroup$
    – Dr Xorile
    Commented Dec 14, 2015 at 4:38
  • $\begingroup$ Yes. What ever polygons you may create by the ruler. $\endgroup$
    – Moti
    Commented Dec 14, 2015 at 4:51
  • $\begingroup$ By "equal in area", do you mean for each polygon individually or for the sum of the 5 polygons to be 1/4 the area of ABCD? $\endgroup$
    – Lawrence
    Commented Dec 14, 2015 at 9:11

1 Answer 1


Where it is clear, I will use XYZ and WXYZ to mean either the shape the area it encloses. Likewise, XY will refer to either the line segment or its length.

enter image description here

Start with the midpoint E of AC.

Since EC = $\frac{1}{2}$AC, we have DEC = $\frac{1}{2}$DAC. (Both triangles have the same perpendicular height from AC to D.) Likewise, BEC = $\frac{1}{2}$BAC.

Summing the areas, we get EBCD = $\frac{1}{2}$ABCD. Call this (1).

Now, DF = FB = $\frac{1}{2}$DB, so DFE = BFE = $\frac{1}{2}$DBE. (All these triangles have the same perpendicular height from DB to E.) Likewise, DFC = BFC = $\frac{1}{2}$DBC. Adding one of each, we get
DEFC = EBCF = $\frac{1}{2}$EBDC. Call this (2).

From (1) and (2), DEFC = EBCF = $\frac{1}{4}$ABCD. With reference to the diagram above, DEFC is composed of the shaded areas P and R, and EBCF is the quadrilateral labelled S. (Note: P, Q and S cross the diagonal lines AC and BD.)

Since DFE = BFE and DFC = BFC, we also have two additional solutions, each composed of two triangles: FEDBCF and FEBDCF. (Although not quadrilaterals, the OP allowed sums and differences of touching triangles as solutions in a comment to the question.)

By a similar construction starting with the midpoint F of DB, we also get ABEF = $\frac{1}{4}$ABCD. ABEF is composed of the shaded areas Q and R. As before, we also get EFACBE and EFCABE as additional solutions.

Now, we have established that P+R = Q+R = S = $\frac{1}{4}$ABCD. So adding the first 3 expressions and decomposing ABCD, we get P+R + Q+R + S = $\frac{3}{4}$(P+Q+R+S+T).

That is, P+Q+R+S+T = 4(T-R), so (T-R) is also $\frac{1}{4}$ABCD.

This gives a total of 8 shapes produced using 5 lines, with each shape occupying an area $\frac{1}{4}$ABCD.

Since we are permitted 9 lines, we have 4 additional lines with which to define acceptable polygons. Here are some approaches that might produce acceptable shapes, possibly by completing the triangle formed from the rays AB and DC. In each case, one vertex is not pinned down, but I leave these here for others to play with.

Since (P+R) + S = $\frac{1}{2}$ABCD, the complement T+Q is also $\frac{1}{2}$ABCD. Since ABE < $\frac{1}{4}$ABCD, there is a point G on AD such that DGE = GABE = $\frac{1}{4}$ABCD. Likewise, there is a point H on AD such that FHA = FHDC = $\frac{1}{4}$ABCD.

Since T > $\frac{1}{4}$ABCD, there is a point I on AB that lies between A and the intersection of AB and the ray DE, such that DAI = DIBE = $\frac{1}{4}$ABCD. Likewise, there is a point J on CD such that AJD = AJCF = $\frac{1}{4}$ABCD.

  • $\begingroup$ Yes, on hold. Also, I think you mean EBCF and not EBFC. $\endgroup$
    – Moti
    Commented Dec 16, 2015 at 5:57
  • $\begingroup$ Try to take advantage of the hint in one of the comments $\endgroup$
    – Moti
    Commented Dec 16, 2015 at 7:28
  • 1
    $\begingroup$ I would suggest you remove this answer, since it discourage others from attempting, since they think it has been answered. $\endgroup$
    – Moti
    Commented Dec 16, 2015 at 18:24
  • $\begingroup$ @Moti I've revised my answer. Based on your response to my query (in comments under your question), this satisfies the revised parameters for the acceptable shapes. $\endgroup$
    – Lawrence
    Commented Dec 26, 2015 at 1:40
  • $\begingroup$ the idea is that the points H and G are found. I would claim that there are many points that could provide polygons with an area 1/4ABCD. In any case, I see only three shapes that satisfy the 1/4ABCD. The solution I suggest is straight clear 5 shapes (polygons) $\endgroup$
    – Moti
    Commented Dec 26, 2015 at 4:44

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