Question

In a triangle $ABC,$ points $D,E$ and $F$ are taken on the sides $BC,CA$ and $AB$ respectively, such that $\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}=n.$ Then Area of $△DEF$ is equal to

• A. $\frac{n^2-n+1}{2(n+1{)}^2}\cdot \text{Area}\left(△ABC\right)$
• B. $\frac{n^2-n+1}{(n+1{)}^2}\cdot \text{Area}\left(△ABC\right)$
• C. $\frac{n^2+n+1}{2(n-1{)}^2}\cdot \text{Area}\left(△ABC\right)$
• D. $\frac{n^2+n+1}{(n-1{)}^2}\cdot \text{Area}\left(△ABC\right)$

Answer 1

The correct option is B $\frac{n^2-n+1}{(n+1{)}^2}\cdot \text{Area}\left(△ABC\right)$

Let $A$ is the origin and the position vector of points $B$ and $C$ are $\overrightarrow{b}$ and $\overrightarrow{c}$ respectively.
As $\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}=n$
By using section formula :
Position vector of $D,E,F$ are $\frac{n\overrightarrow{c}+\overrightarrow{b}}{n+1},\frac{\overrightarrow{c}}{n+1},\frac{n\overrightarrow{b}}{n+1}$ are respectively.
So, $\overrightarrow{ED}=\overrightarrow{AD}-\overrightarrow{AE}=\frac{(n-1)\overrightarrow{c}+\overrightarrow{b}}{n+1}$ and
$\overrightarrow{EF}=\frac{n\overrightarrow{b}-\overrightarrow{c}}{n+1}$
Now, vector area of triangle $DEF=\frac{1}{2}(\overrightarrow{EF}\times \overrightarrow{ED})$
$=\frac{1}{2(n+1{)}^2}\left[\right(n\overrightarrow{b}-\overrightarrow{c})\times \{(n-1)\overrightarrow{c}+\overrightarrow{b}\left\}\right]=\frac{1}{2(n+1{)}^2}\left[\right(n^2-n)\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{c}]=\frac{n^2-n+1}{2(n+1{)}^2}(\overrightarrow{b}\times \overrightarrow{c})=\frac{n^2-n+1}{(n+1{)}^2}\cdot \text{Area}\left(△ABC\right)$