Question

In triangle $△ABC$,D, E, F are points of trisection of BC, AC and AB respectively. Which of the following statements is not true ?

• A. $Area△EDC=2/9area△ABC$
• B. $Area△FBD=1/8areaAFDC$
• C. $Area△DEF=2/9area△ABC$
• D. $Area(△EDC+△DBF+△AFE)=2Area△DEF$

Answer 1

The correct option is D $Area(△EDC+△DBF+△AFE)=2Area△DEF$

Let $AB,BC,AC$ be $3x,3y,3z$
So, $DF,BD,EC$ be $x,y,z$
$\frac{A\left(\Delta EDC\right)}{A\left(\Delta ABC\right)}=\frac{\frac{1}{2}\times EC\times CDsinC}{\frac{1}{2}\times AC\times BCsinC}$
$\frac{A\left(\Delta EDC\right)}{A\left(\Delta ABC\right)}=\frac{\frac{1}{2}\times z\times 2ysinC}{\frac{1}{2}\times 3z\times 3ysinC}$
$\frac{A\left(\Delta EDC\right)}{A\left(\Delta ABC\right)}=\frac{2}{9}$
Hence, option $A$ is correct.
$\frac{A\left(\Delta ABC\right)}{A\left(\Delta BFD\right)}=\frac{\frac{1}{2}\times AB\times BCsinB}{\frac{1}{2}\times BF\times BDsinB}$
$\frac{A\left(\Delta ABC\right)}{A\left(\Delta BFD\right)}=\frac{\frac{1}{2}\times 3x\times 3ysinB}{\frac{1}{2}\times x\times ysinB}$
$\frac{A\left(\Delta ABC\right)}{A\left(\Delta BFD\right)}=\frac{9}{1}$
$\frac{A\left(\Delta ABC\right)}{A\left(\Delta BFD\right)}-1=\frac{9}{1}-1$
$\frac{A\left(AFDC\right)}{A\left(\Delta BFD\right)}=\frac{8}{1}$
$A\left(\Delta BFD\right)=\frac{1}{8}\times A\left(AFDC\right)$
Hence, option $B$ is also correct.
Same can be proved for option $C$ as in $A$.
$A\left(\Delta EDC\right)+A\left(\Delta DBF\right)+A\left(\Delta AFE\right)=2A\left(\Delta DEF\right)$
$A\left(\Delta ABC\right)-A\left(\Delta DEF\right)=2A\left(\Delta DEF\right)$
$A\left(\Delta ABC\right)=3A\left(\Delta DEF\right)$ which is incorrect since $A\left(\Delta ABC\right)=\frac{9}{2}A\left(\Delta DEF\right)$