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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.
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.
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.
