RATIO WITH AREA

[MATHPRITOM]

ABCDEF be a hexagon. $ A_1,B_1,C_1,D_1,E_1,F_1 $ are the midpoints on AB,BC,CD,DE,EF,FA respectively.If $ (ACE) =\frac{1}{4}(ABCDEF)$, Prove that ,$(A_1BB_1)+(C_1DD_1)+(E_1FF_1)=\frac{3}{16}(ABCDEF) $ .Here, (x) indicates the area of x.

[Raiyan Jamil]

Triangle $\frac 14(ACE)=(ABCDEF)$. Now, as ABC+CDE+EFA+ACE=ABCDEF, so 3*ACE=ABC+CDE+EFA. Therefore,
$(ABC)+(CDE)+(EFA)=\frac 34(ABCDEF)$. Now, it is said that, $A_1;B_1;C_1;D_1;E_1;F_1$ are the midpoints on AB,BC,CD,DE,EF,FA respectively. So,
$\frac 14(ABC)=(A_1BB_1)$;
$\frac 14(CDE)=(C_1DD_1)$;
$\frac 14(EFA)=(E_1FF_1)$;
and so,$ \frac 14(ABC+CDE+EFA)=(A_1BB_1)+(C_1DD_1)+(E_1FF_1)$
and then, $(A_1BB_1)+(C_1DD_1)+(E_1FF_1)=\frac 14 \times \frac 34(ABCDEF)$
or, $(A_1BB_1)+(C_1DD_1)+(E_1FF_1)=\frac 3{16}(ABCDEF)$

[tanmoy]

Triangle $\frac 14(ACE)=(ABCDEF)$.

$\frac{1}{4}(ACE)\neq (ABCDEF).
\frac{1}{4}(ABCDEF)=(ACE)$

[Raiyan Jamil]

Sorry, that was a silly mistake.