Prove that $2A+A_0=A_1+A_2+A_3$,where $A$ is the area of the triangle $ABC.$

Question

If $A_0$ denotes the area of the triangle formed by joining the points of contact of the inscribed circle of the triangle $ABC$ and the sides of the triangle;$A_1,A_2,A_3$ are the corresponding areas for the triangles thus formed with the escribed circles of the triangle $ABC.$Prove that $2A+A_0=A_1+A_2+A_3$,where $A$ is the area of the triangle $ABC.$

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I found the area $A_0=A-\frac{1}{2}(s-a)^2-\frac{1}{2}(s-b)^2-\frac{1}{2}(s-c)^2$

Is my $A_0$ correct?
I am not able to find $A_1,A_2,A_3$.Please help.

Answer 1

Is my $A_0$ correct?

No, it is not correct. Let $D,E,F$ be the tangent point of the inscribed circle with the side $BC,CA,AB$ respectively. Then, noting that $$AE=AF=s-a,\quad BD=BF=s-b,\quad CE=CD=s-c$$ where $s=(a+b+c)/2$, we have $$\begin{align}A_0&=[\triangle{ABC}]-([\triangle{AEF}]+[\triangle{BDF}]+[\triangle{CDE}])\\&=A-\frac 12(s-a)^2\sin A-\frac 12(s-b)^2\sin B-\frac 12(s-c)^2\sin C\end{align}\tag1$$

By the way, let $K,L,M$ be the tangent point of the escribed circle in $\angle A$ with the side $BC,CA,AB$ respectively. Then, noting that $$CK=CL=s-b,\quad BK=BM=s-c,$$ we have $$\begin{align}[\triangle{KLM}]&=[\triangle{AML}]-([\triangle{ABC}]+[\triangle{BKM}]+[\triangle{CKL}])\\&=\frac 12s^2\sin A-A-\frac 12(s-c)^2\sin(\pi-B)-\frac 12(s-b)^2\sin(\pi-C)\\&=\frac 12s^2\sin A-A-\frac 12(s-c)^2\sin B-\frac 12(s-b)^2\sin C\end{align}$$ Similarly, the areas of the other triangles formed with escribed circles are given by $$\frac 12s^2\sin B-A-\frac 12(s-a)^2\sin C-\frac 12(s-c)^2\sin A$$ $$\frac 12s^2\sin C-A-\frac 12(s-a)^2\sin B-\frac 12(s-b)^2\sin A$$

Hence, from $(1)$, we have $$2A+A_0-A_1-A_2-A_3$$$$=2A+A-\frac 12(s-a)^2\sin A-\frac 12(s-b)^2\sin B-\frac 12(s-c)^2\sin C-(\frac 12s^2\sin A-A-\frac 12(s-c)^2\sin B-\frac 12(s-b)^2\sin C)-(\frac 12s^2\sin B-A-\frac 12(s-a)^2\sin C-\frac 12(s-c)^2\sin A)-(\frac 12s^2\sin C-A-\frac 12(s-a)^2\sin B-\frac 12(s-b)^2\sin A)$$$$=6A+\frac 12\sin A(-(s-a)^2-s^2+(s-c)^2+(s-b)^2)$$$$+\frac 12\sin B(-(s-b)^2+(s-c)^2-s^2+(s-a)^2)$$$$+\frac 12\sin C(-(s-c)^2+(s-b)^2+(s-a)^2-s^2)$$$$=6A+\frac 12\sin A(-2bc)+\frac 12\sin B(-2ca)+\frac 12\sin C(-2ab)=6A-2A-2A-2A=0$$

Answer 2

Hint:

$A_1=\dfrac{A^2}{2R(s-a)}$ and $A_{o}=\dfrac{A^2}{2Rs}$

Plug in the corresponding values in \begin{align} A_1+A_2+A_3-A_{o}&=\dfrac{A^2}{2R}\cdot\bigg[\dfrac{1}{s-a}+\dfrac{1}{s-b}+\dfrac{1}{s-c}-\dfrac{1}{s}\bigg]\\&=\dfrac{A^2}{2R}\cdot\dfrac{abc}{A^2}\\&=\dfrac{2A^2\cdot abc}{4R\cdot A^2}\\&=2A\ \ \ \ \because{A=\frac{abc}{4R}}\end{align}