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

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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 $A_0=A-frac12(s-a)^2-frac12(s-b)^2-frac12(s-c)^2$


Is my $A_0$ correct?

I am not able to find $A_1,A_2,A_3$.Please help.







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  • 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














up vote
0
down vote

favorite












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 $A_0=A-frac12(s-a)^2-frac12(s-b)^2-frac12(s-c)^2$


Is my $A_0$ correct?

I am not able to find $A_1,A_2,A_3$.Please help.







share|cite|improve this question




















  • 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












up vote
0
down vote

favorite









up vote
0
down vote

favorite











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 $A_0=A-frac12(s-a)^2-frac12(s-b)^2-frac12(s-c)^2$


Is my $A_0$ correct?

I am not able to find $A_1,A_2,A_3$.Please help.







share|cite|improve this 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 $ABC.$




I found the area $A_0=A-frac12(s-a)^2-frac12(s-b)^2-frac12(s-c)^2$


Is my $A_0$ correct?

I am not able to find $A_1,A_2,A_3$.Please help.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 21 '16 at 8:27









Vinod Kumar Punia

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  • 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
















  • 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















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




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




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










2 Answers
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1
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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
$$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$



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
$$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
Similarly, the areas of the other triangles formed with escribed circles are given by
$$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
$$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$



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






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  • I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
    – Vinod Kumar Punia
    Apr 21 '16 at 10:52






  • 1




    @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






  • 1




    @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

















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0
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Hint:



$A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$




Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign






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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    up vote
    1
    down vote



    accepted











    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
    $$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$



    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
    $$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
    Similarly, the areas of the other triangles formed with escribed circles are given by
    $$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
    $$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$



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






    share|cite|improve this answer






















    • I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
      – Vinod Kumar Punia
      Apr 21 '16 at 10:52






    • 1




      @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






    • 1




      @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














    up vote
    1
    down vote



    accepted











    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
    $$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$



    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
    $$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
    Similarly, the areas of the other triangles formed with escribed circles are given by
    $$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
    $$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$



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






    share|cite|improve this answer






















    • I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
      – Vinod Kumar Punia
      Apr 21 '16 at 10:52






    • 1




      @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






    • 1




      @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












    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted







    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
    $$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$



    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
    $$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
    Similarly, the areas of the other triangles formed with escribed circles are given by
    $$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
    $$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$



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






    share|cite|improve this answer















    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
    $$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$



    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
    $$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
    Similarly, the areas of the other triangles formed with escribed circles are given by
    $$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
    $$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$



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







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    edited Apr 21 '16 at 10:45

























    answered Apr 21 '16 at 9:19









    mathlove

    87.1k877208




    87.1k877208











    • I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
      – Vinod Kumar Punia
      Apr 21 '16 at 10:52






    • 1




      @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






    • 1




      @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
















    • I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
      – Vinod Kumar Punia
      Apr 21 '16 at 10:52






    • 1




      @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






    • 1




      @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















    I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
    – Vinod Kumar Punia
    Apr 21 '16 at 10:52




    I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
    – Vinod Kumar Punia
    Apr 21 '16 at 10:52




    1




    1




    @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





    @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




    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




    1




    1




    @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




    @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




    I got it,thank you very much.@mathlove
    – Vinod Kumar Punia
    Apr 21 '16 at 12:19










    up vote
    0
    down vote














    Hint:



    $A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$




    Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign






    share|cite|improve this answer


























      up vote
      0
      down vote














      Hint:



      $A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$




      Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign






      share|cite|improve this answer
























        up vote
        0
        down vote










        up vote
        0
        down vote










        Hint:



        $A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$




        Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign






        share|cite|improve this answer















        Hint:



        $A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$




        Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 17 at 4:39

























        answered Aug 17 at 4:24









        mnulb

        1,325620




        1,325620






















             

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