Let $M$, $N$, $L$ be midpoints of $AB$, $BC$, $AD$. If $CL\parallel AB$ and $NQ\parallel ML$, find the area of $ABCD$.

Question

Let $M$, $N$, and $L$ be the midpoints of $AB$, $BC$ and $AD$, respectively. Suppose that $CL\parallel AB$. A point $Q$ on $CL$ is such that $NQ\parallel ML$. Calculate the area of ​the quadrangular region $ABCD$ as a function of areas $\Bbb A$, $\Bbb B$, $\Bbb C$.

I tried in so many ways but I couldn' isolate this area in terms of $\Bbb A$, $\Bbb B$, and $\Bbb C$, because I coudn't list the area [CLD] with the areas $\Bbb A$, $\Bbb B$, and $\Bbb C$.

The answer is $$[ABCD]=8(\Bbb A+\Bbb B-\Bbb C).$$ I tried to draw AN and divide the areas A and B in $A_1,A_2,B_1,B_2$, and I got the equations: $$A_2+B_1+H=D,$$ $$A_1+B_2+F=C+G+E.$$ How can I do this?

Answer 1

Let $P=ML\cap AN$, $E=CL\cap DB$, $F=AN\cap BD$, and $R=AE\cap PL$. Because $AB\parallel CL$ and $L$ biescts $AD$, $E$ is the midpoint of $BC$. Since $M$ is a midpoint of $AB$, $ME\parallel AD$. Therefore, $\square ALEM$ is a parallelogram. Therefore, the diagonals $AE$ and $ML$ bisect each others. Hence, $R$ is a midpoint of $AE$. As $N$ is a midpoint of $BC$ and $AB\parallel CL$, we see that $NR\parallel AB$. Therefore, $$[MNA]=[MRA].$$ This shows that $$\Bbb C=[MPN]=[MNA]-[MPA]=[MRA]-[MPA]=[APR].$$

We then have $$[ARL]=[APL]-[APR]=\Bbb A-\Bbb C.$$ Note that $\triangle ARL\sim\triangle AED$ because $BD\parallel ML$ ($M$ and $L$ are midpoints of $AB$ and $AD$). Therefore, $$\frac{[ARL]}{[AED]}=\left(\frac{AL}{AD}\right)^2=\frac14.$$ Since $ED=EB$, we get $$[AEB]=[AED].$$ So $$[ABD]=[AEB]+[AED]=2[AED]=2\cdot 4[ARL]=8(\Bbb A-\Bbb C).$$

Note that $\triangle CNQ\sim \triangle CBE$ because $BD\parallel ML$ and $ML\parallel NQ$, so $$\frac{[CNQ]}{[CBE]}=\left(\frac{CN}{CB}\right)^2=\frac14.$$ Since $L$ bisects $AD$ and $CL\parallel AB$, $E$ also bisects $BD$. Therefore $$[CBE]=[CED].$$ Hence $$[BCD]=[CBE]+[CED]=2[CBE]=2\cdot 4[CNQ]=8\Bbb B.$$ That is, $$[ABCD]=[ABD]+[BCD]=8(\Bbb A-\Bbb C)+8\Bbb B=8(\Bbb A+\Bbb B-\Bbb C).$$

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If all else fails, you can use assign coordinates to the problem. WLOG, let $A=(0,0)$, $L=(2,0)$, $D=(4,0)$, $M=(0,1)$, and $B(0,2)$. We can do this because area ratios are preserved under affine transformations.

If $C=(2,2c)$, then $N=(1,c+1)$. Since $NQ\parallel ML$, $NQ$ is given by the linear equation $$y-c-1=-\frac{1}{2}(x-1).$$ So $Q=\left(2,c+\frac12\right)$. The line $ML$ is $$y=-\frac{1}{2}(x-2)$$ and the line $AN$ is $$y=(c+1)x.$$ If $P=AN\cap ML$, then $$P=\left(\frac{2}{2c+3},\frac{2c+2}{2c+3}\right).$$

We have $$[APL]=\frac{1}{2}\cdot\frac{2c+2}{2c+3}\cdot 2=\frac{2c+2}{2c+3},$$ $$[CNQ]=\frac{1}{2}\cdot 1\cdot\left(2c-c-\frac12\right)=\frac{2c-1}{4},$$ and $$[MPN]=\frac12\left|\begin{array}{ccc} 1&0&1\\ 1&\frac{2}{2c+3}&\frac{2c+2}{2c+3}\\ 1&1&c+1 \end{array}\right|=\frac{1}{2}\left(\frac{2c+1}{2c+3}\right).$$ This means $2[APL]-2[MPN]=1$ and $c=2[CNQ]+\frac12$. However, $$[ABCD]=\frac{1}{2}\cdot 2\cdot (2+2c)+\frac12\cdot 2\cdot 2c=2+2c+2c=4c+2.$$ That is $$[ABCD]=4\left(2[CNQ]+\frac12\right)+2=8[CNQ]+4=8[CNQ]+4\big(2[APL]-2[MPN]\big).$$ Hence $$[ABCD]=8[APL]+8[CNQ]-8[MPN]=8\Bbb A+8\Bbb B-8\Bbb C.$$

Answer 2

Angles $BAD$ and $ABC$ are arbitraries so if the required relation is non constant then the problem is not determined. Consequently in order to verify the relation $[ABCD]=8(\Bbb A+\Bbb B-\Bbb C)$ we choose a convenient quadrilateral for easy calculation.

Let the trapezoid $ABCD$ where $A=(0,0),\space B=(0,2a)\space C=(a,2a)\space D=((2a,0)$ We have $M=(0,a),\space L=(a,0)\space N=(\dfrac a2,2a)$ and $CL\parallel AB$.

$AN$ and $ML$ determine a point of intersection $R=(\dfrac a5,\dfrac{4a}{5})$ and the point $Q=(a,\dfrac {3a}{2})$ is given by intersection of $CL$ and $NQ$.

With this triangles $NQC$ and $ARL$ clearly have an area of $\dfrac{a^2}{8}$ and $\dfrac{2a^2}{5}$ respectively and usually calculation gives the other shadow triangle $MNR$ has an area equal to $\dfrac{3a^2}{20}$.

Now $$8( \dfrac{2a^2}{5}+\dfrac{a^2}{8}-\dfrac{3a^2}{20})=\dfrac{8\cdot15a^2}{40}=3a^2$$

This $3a^2$ is precisely the area of the trapezoid.

Answer 3

Given the midpoints M, L and N, we have ML || BD || NQ and ML = BT = $2$NQ. Construct NP || AB || CL to have the similar $\triangle$AME and $\triangle$NEP, as well as all the equal segment lengths $x$, $y$ and $z$ indicated in the diagram. Then,

$$\frac{EM}{x-EM}= \frac y{y+z} \implies \frac{EM}x = \frac{y}{2y+z},\>\>\>\frac{EL}x= \frac{3y+2z}{2y+z}$$

Let [.] denote areas and evaluate $\Bbb A$, $\Bbb B$ and $\Bbb C$ in terms of $[ABD]$ and $[CBD]$

\begin{align} \Bbb A & = \frac{EL}{ML} [AML] =\frac{EL}{2x} \cdot\frac14[ABD] = \frac{3y+2z}{8(2y+z)} [ABD] \\ \Bbb B & = \frac14 [CBT] = \frac18 [CBD] \\ \Bbb C & = \frac{EM}{PM} [NMP] = \frac{EM}{x}\cdot\frac12[NPLQ] = \frac{EM}{2x}([NGTQ]+[PLTG]) \\ &\hspace{0cm}=\frac{y}{2(2y+z)}(\frac14[CBD]+\frac14[ABD]) =\frac{y}{8(2y+z)} ([CBD]+[ABD]) \end{align}

Then

$$\Bbb A + \Bbb B - \Bbb C = \frac{y+z}{8(2y+z)} (2[ABD] + [CBD] ) $$

Recognize $\frac yz = \frac{[ABD]}{[CBD]}$ and substitute it into above equation to obtain

$$\Bbb A + \Bbb B - \Bbb C = \frac18 ([ABD] + [CBD] ) = \frac18 [ABCD] $$