Making the area of the quadrilateral and the area of a triangle the same

Question

$ABCD$ is a quadrilateral and $X$ is a given point on AD. Find a point Y in AB such that the area of the $\triangle AXY$ is equal to that of $ABCD$. Hence show how to divide the quadrilateral $ABCD$ into three equal parts by straight lines drawn through $X$.

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I cannot for the life of me solve the first problem. No matter what I try, what parallel lines I draw, I cannot make a triangle which has the same area of the triangle in this specific configuration. If the point $Y$ was opposite to point $X$, i.e, if it was on $BC$, then I can solve. But I just am unable to transfer this intuition to the case with $AB$. And also I fail to see how the first part is related to the 2nd part of the question. Could anyone explain it to me?

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This is how I solved the 2nd part of the problem.

I joined $BX$ and $CX$. $AE$ and $DF$ are drawn parallel to $BX$ and $CX$ respectively. Then I join $XE$ and $XF$. The triangle $XEF$ has the same area as the quadrilateral $ABCD$. And we divide $EF$ in the ratio $\frac{1}{3}$. This point $P$ divides the quadrilateral into two regions having area ratio $1:2$. Making another line is essentially the same process.

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Answer 1

Even if not explicitly stated, I'll assume quadrilateral $ABCD$ is convex.

Draw from $C$ the parallel $CC'$ to $BD$, intersecting $AB$ at $C'$. Triangle $ADC'$ has the same area as $ABCD$ (because triangles $BDC$ and $BDC'$ have equal altitudes with respect to common base $BD$).

Draw from $D$ the parallel $DY$ to $CX$, intersecting $AB$ at $Y$. Triangle $AXY$ has the same area as $ADC'$ (because triangles $YXC'$ and $DXC'$ have equal altitudes with respect to common base $XC'$).

Hence $AXY$ has the same area as $ABCD$.

EDIT.

To divide quadrilateral $ABCD$ into three polygons of equal area with three lines drawn through $X$ one can then proceed as follows.

Draw point $P$ on $AY$ such that $AP={1\over3}AY$: the area of triangle $APX$ is then ${1\over3}$ the area of $ABCD$ and if $P$ lies on side $AB$ then $APX$ the first desired first polygon.

If $P$ is outside $AB$ (as in figure below) then $PX$ cuts $BC$ at some point $E$, but the area of $ABEX$ is less than the desired area, by an amount equal to the area of triangle $BPE$. But we can easily construct a triangle $QEX$ with the same area as $BPE$: just construct on $EC$ point $Q$ such that $$ QE:BE=PE:XE. $$ Quadrilateral $ABQX$ is then the desired first polygon.

To construct a second polygon we can repeat everything on line $DC$: first of all construct $Y'$ on line $DC$ such that triangle $DXY'$ has the same area as $ABCD$, then divide $DY'$ into three parts, to find $P'$ (and $Q'$ if needed).

The third polygon is what is left of $ABCD$ by the other two.

Of course I assumed $Q$ to lie on side $BC$, but that might not be the case: in that event we should modify $ABQX$ to a pentagon, following the same path discussed above.

Answer 2

$\underline{\mathrm{Part\space I}}:\space \mathrm{Determination\space of\space point\space }\mathbf\it{Y}$ Our construction has two steps. In the first step, a triangle is found, which has the area of the given quadrilateral and shares one of the sides with it. In the second step, the sought triangle is constructed, so that it has the area of the triangle found in the previous step. Please note that we decided to leave the proof of the construction up to OP.

$ABCD$ shown in $\mathrm{Fig.\space 1}$ is a quadrilateral, the four sides of which were chosen arbitrarily, and $X$ is any point lying on its side $AB$. First, a line is drawn to join the opposite vertices $A$ and $C$ of $ABCD$. Next, a line parallel to $AC$ is constructed through the vertex $D$ to intersect the extended side $BC$ at $E$. Now we have the triangle $ABE$ which has the area of the quadrilateral $ABCD$ and the two figures share the side $AB$. This is a pretty standard construction and can be found in any good textbooks on geometry.

$\mathrm{Fig.\space 2}$ illustrates how to obtain the desired triangle with the area of the triangle $ABE$ and has $AX$ as one of its side. To begin the construction, draw two perpendicular lines to $AB$ through $X$ and $B$. Next, a parallel line to $AB$ is drawn through $E$, one of the vertices of the triangle $ABE$, to intersect the perpendiculars through $X$ at $F$. After joining $A$ and $F$, extend it to meet the perpendiculars through $B$ at $H$. Finally, complete the construction by drawing a line parallel to $AB$ through $H$ to cut the extended side $AD$ of the quadrilateral $ABCD$ at $Y$. Since the triangle $AXY$ has the area of $ABE$, both $ABCD$ and $AXY$ have equal areas.

We have noticed that, as $X$ moves from $A$ to $B$, $Y$ travels from $\infty$ to $Z$ (see $\mathrm{Fig.\space 3}$). $$ $$ $\underline{\mathrm{Part\space II}}:\space \mathrm{Partition\space of\space a\space quadrilateral\space into\space three\space equal\space parts\space}$ In this 3-step-construction described below, we try to show how the first part is related to the division of the quadrilateral. This, we do by exclusive use of the triangle $AXY$.

First, we construct the point of concurrence of the medians $G$ known as the centroid of $AXY$. As shown in $\mathrm{Fig.\space 4}$, three sub-triangles $AXG$, $XYG$, and $YAG$, which are formed by the lines joining $G$ to the vertices of $AXY$, have the same area. Therefore, the area of each of the sought subdivisions of the quadrilateral is also equal to the area of a sub-triangle. We make use of this fact to obtain one of the subdivisions, namely $AXP$, by drawing a line parallel to $AB$ through $G$ to cut $AD$ at $P$. It is evident that the areas of the triangles $AXG$ and $AXP$ are the same, because they both have the same base and height.

Next, as shown in $\mathrm{Fig.\space 5}$, we draw a line parallel to $DG$ through $Y$ to cut $CD$ at $K$. To obtain the second and the third subdivisions, two lines are drawn to join $G$ to $K$ and $X$. The area of $PXG$ is equal to that of $PAG$ , because they both have the same base and height. In similar manner, the areas of the two triangles $DGK$ and $DGY$ are equal. Since the area $PGD$ is common to both the triangle $YAG$ and the pentagon $PXGKD$, their areas are equal. Therefore, the remaining pentagonal subdivision $XBCKG$ has the same area of either $AXP$ or $PXGKD$.

Only two of the three lines $PX$, $GX$, and $GK$ drawn to partition the quadrilateral goes through $X$. In order to rectify this problem, as shown in $\mathrm{Fig.\space 6}$, we need to continue the construction by drawing a line parallel to $KX$ through $G$ to cut $CD$ at $Q$. By doing this, we exchanged two equal areas $GXM$ (which was a part of the pentagon $XBCKG$) and $KQM$ (which was a part of the pentagon $PXGKD$) to obtain what we want.

It should be noted that there is no unique construction protocol capable of partitioning all quadrilateral to satisfy the given requirements. Therefore, we have to devise a construction as demanded by the case at hand.