Beautiful triangle problem

Let's label the figure as above. We note that external angle $\angle DEC = \blue{\angle BDE} + \red{\angle DBE}$ . Since $\angle DEC = \angle DEF + \angle FEC$ , and $\angle DEF = \blue{\angle BDE}$ $\implies \angle FEC = \red{\angle DBE}$ . Therefore $\triangle DBE$ and $\triangle FEC$ are similar. We also note that $\triangle ABC$ is also similar to the two small triangle. Since the area of similar triangles are proportional to the square of their linear dimension. Then, if $\triangle DBE$ has a base length of $2a$ , $\triangle FEC$ has a base length of $3a$ . Therefore, the base length of $\triangle ABC$ is $5a$ and the area $[ABC] = 25$ , and the area of the yellow quadrilateral $[ADEF] = [ABC]-[DBE]-[FEC] = 25 - 4 - 9 = \boxed{12}$ .

let $[BD]||[FH]$

$\angle FBG = \angle BGD$ Hence $[BF]||[DG]$

so $\angle BFG = \angle EGH$ , $\angle BGF = \angle EHG$ hence $BFG$ and $EGH$ are similar and $[AE]||[BG]$

their areas ratio is $\dfrac{4}{9}$ so their similarity ratio is $\dfrac{2}{3}$

let $|FG|=2a$ then $|GH|=3a$

heights of $BFG$ and $DGH$ are equal so their area ratio will be $\dfrac{2}{3}$

area of $BFG$ is $4$ then area of $DGH$ is $6$ and area of $EDH$ is $3$ . So $\dfrac{|DG|}{|ED|}=\dfrac{6}{3}=2=\dfrac{ar(BDG)}{ar(EDB)}$

$BDGF$ is a parallelogram hence $ar(BDG)=ar(BFG)=4$ so area of $EBD$ is $2$

again $ABGE$ is a parallelogram Hence $ar(ABE)=ar(BEG)=4+2=6$

area of $ABGE$ is sum of the areas of $ABE$ and $BEG$ $\Rightarrow 6+6=12$

Note:sorry i couldnt draw DH good

Let the vertices of the whole triangle be $A, B, C$ , the vertex of the yellow quadrilateral on the side $\overline {BC}$ be $D$ , on the side $\overline {CA}$ be $E$ , and on the side $\overline {AB}$ be $F$ . Let $|\overline {BC}|=a, |\overline {CA}|=b, |\overline {AB}|=c$ . Since $\overline {DF}$ and $\overline {CA}$ are parallel, therefore, let $|\overline {BD}|=pa, |\overline {DC}|=qa, |\overline {BF}|=pc, |\overline {FA}|=qc$ . Since $\overline {DE}$ and $\overline {BA}$ are parallel, therefore $|\overline {CE}|=qb, |\overline {EA}|=pb$ . Area of $\triangle {BFD}=4=\dfrac{1}{2}pa\times {pc}\times {\sin B}$ , where $B=\angle {ABC}$ . So $p^2\triangle =4$ , where $\triangle$ is the area of $\triangle {ABC}=\dfrac{1}{2}ac\sin B=\dfrac{1}{2}bc\sin A=\dfrac{1}{2}ab\sin C$ . ( $A=\angle {BAC}, C=\angle {ACB}$ ). Therefore $\triangle =\dfrac{4}{p^2}$ . Area of $\triangle {DCE}=9=\dfrac{1}{2}qa\times {qb}\times {\sin C}=q^2\triangle$ . So $\triangle =\dfrac{9}{q^2}$ . From this we get $q=\dfrac{3p}{2}$ . Hence, area of the quadrilateral $AEDF$ is $pb\times {qc}\times {\sin C}=2pq\triangle =3p^2\triangle =3\times 4=\boxed {12}$