Let $E , G , F , H$ be the midpoints of sides $AB , BC , CD , DA ,$ respectively of quadrilateral $ABCD.$ The common points of segments $AF , BH,CE,DG $ divide
each of into three parts, as shown in the figure.
Given that $$\frac{LK}{GD}=\frac{LI}{AF}=\frac{IJ}{BH}=\frac{JK}{CE}=k$$
Find $K.$ And also construct the $ABCD$ such that $ABCD$ is a parallelogram and still satisfies the condition. Additionally, prove that $LIJK$ is a trapezoid.
The answer is $2/5.$ And I want an elementary proof. The question's subpart was asked here and We think a separate new question should be there to discuss about the elementary method.
Note that $EFGH$ is a parallelogram. And $Ar[ABCD]=2Ar[EFGH].$ And define $M=AC\cap BD.$ Note that $$2k^2\times Ar[MIJ]=2 Ar[MHB]=2Ar[MEB]=Ar[MAB].$$
Hence, we have $2k^2\times Ar[IJKL]=Ar[ABCD]\implies 2k^2\times Ar[IJKL]=Ar[EFGH].$
Best Answer
I will try to spend as few words as possible. The figure should be self-explanatory.
First of all notice as basic fact $$ [ADE]+[CBG]=[BAF]+[DCH]=\frac12[ABCD], $$ which implies $$ [AFCH]=[BGDE]=\frac12[ABCD]. $$
Let $M$ be the intersection of the bimedians $FH$ and $EG$. As well known and can be trivially proven $M$ bisects both bimedians.
Consider the quadrilateral $BGDE$ (first figure). Since $M$ is the midpoint of $EG$ we have: $$ \begin{cases} [BMG]=[BME]\\ [DME]=[DMG] \end{cases}\implies [BMG]+[DME]=\frac12[BGDE]=\frac14[ABCD], $$ which implies: $$ [IMJ]+[LMK]=\frac k4[ABCD]. $$ Applying the same argument to the quadrilateral $AFCH$ and summing the results one obtains: $$ [IJKL]=\frac k2[ABCD].\tag1 $$
The second figure tells us: $$ [ALE]+[BIF]+[CJG]+[DKH]=[IJKL]\tag2 $$
From the third figure we have: $$ [ABF]=2[ALE]+[BIF]+k[ABF]\implies (1-k)[ABF]=2[ALE]+[BIF]\tag3 $$
Summing the equation $(3)$ for all four triangles $ABF,BCG,CDH,DAE,$ one finally obtains: $$ (1-k)[ABCD]=3[IJKL]=\frac{3k}2[ABCD]\implies k=\frac25.\tag4 $$