Angle bisector related question

Question

The internal bisectors of the angles of a triangle ABC meet the sides in D,E,and F.Show that area of the triangle DEF is equal to $\frac{2\Delta\times abc}{(b+c)(c+a)(a+b)}$,here $\Delta $is area of triangle ABC

If I choose B as origin,C as$(a,0)$,$a$ is side length BC.what should i take coordinates of A,can i get answer with this approach or any better approach?

Answer 1

Let $D,E,F$ be on $BC,CA,AB$ respectively.

Now, since we know that $$BD:DC=c:b,\quad AF:FB=b:a$$ the area of $\triangle{BDF}$ is $$\Delta\times \frac{c}{b+c}\times \frac{a}{a+b}=\frac{ac}{(a+b)(b+c)}\Delta.$$ Similarly, we can have $$[\triangle{CED}]=\frac{ab}{(c+a)(b+c)}\Delta,\quad [\triangle{AEF}]=\frac{bc}{(a+b)(c+a)}\Delta.$$

Thus, the area of $\triangle{DEF}$ is $$[\triangle{ABC}]-[\triangle{BDF}]-[\triangle{CED}]-[\triangle{AEF}]=\frac{2abc}{(a+b)(b+c)(c+a)}\Delta.$$

Answer 2

Hint: If $D\in BC$, $E\in CA$, and $F\in AB$, then $\frac{AB}{BC}=\frac{AD}{DC}$. We have similar conditions for $E$ and $F$. Then, calculate $\frac{[AEF]}{[ABC]}$, $\frac{[BFD]}{[ABC]}$, and $\frac{[CDE]}{[ABC]}$, where $[P]$ is the area of a geometric object $P$.