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

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 $ABC.$

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.

• can you be more explicit on how you define the "triangles thus formed with the escribed circles...". Is not clear... – Chip Apr 21 '16 at 8:30
• The escribed circle opposite vertex $A$ touches $AB$ produced,$AC$ produced and $BC$,so $A_1$ is the area of triangle formed by points of contact of escribed circle with $BC,AB$ produced,$AC$ produced.Same for escribed circles opposite $B$ and $C.$ – Vinod Kumar Punia Apr 21 '16 at 8:34

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

• I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove – Vinod Kumar Punia Apr 21 '16 at 10:52
• @VinodKumarPunia: Since $a=BC=BK+KC=BM+CL$, we have $a+b+c=b+c+BM+CL=(b+CL)+(c+BM)=AL+AM$, and so we have $AL=AM=s$, from which $CK=CL=AL-AC=s-b$ and $BK=BM=AM-AB=s-c$ follow. – mathlove Apr 21 '16 at 11:11
• How can we prove that $AE=AF=s-a,BD=BF=s-b,CD=CE=s-c$?@mathlove – Vinod Kumar Punia Apr 21 '16 at 11:18
• @VinodKumarPunia: Let $AE=AF=x,BD=BF=y,CD=CE=z$. Solving $z+y=a,x+z=b,x+y=c$ will give you the answer. – mathlove Apr 21 '16 at 11:24
• I got it,thank you very much.@mathlove – Vinod Kumar Punia Apr 21 '16 at 12:19

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}