Properties of Triangles

nobody tryin this un????

soln of Δ..

i am weak in that... [3]

and upar se.. exradii...

yes gr8 idea 4 objectives...

Its verry long................I would be using some things as such(the bold ones)........if u want i will explain later.....

(r1+r2+r3-r)²=r1²+r2²+r3²+r²-2r(r1+r2+r3)+2(r1r2+r2r3+r3r1)

We know that r1+r2+r3-r=4R

and r1r2+r2r3+r3r1=s²

=> (4R)²=r1²+r2²+r3²+r²-2r(r1+r2+r3)+2s²

=>r1²+r2²+r3²+r²=16R²+2(rr1+rr2+rr3)-2s² ----------------(1)

Also, 2(rr1+rr2+rr3)=2[S²/s(s-a) +S²/s(s-b)
+S²/s(s-c)]
=>2S² [(s-b)(s-c)+(s-c)(s-a)+(s-a)(s-b)]
=>2[3s²-s(b+c+c+a+a+b)+bc+ca+ab]
=>2[3s²-s.4s+bc+ca+ab]
=> -2s²+2(ab+bc+ca)

From(1)
r1²+r2²+r3²+r²=16R²-2s²+2(ab+bc+ca)-2s²
=>16R²-[(a+b+c)²-2(ab+bc+ca)]
=>16R²-(a²+b²+c²]

=>r1²+r2²+r3²+r²+a²+b²+c²+R²=17R²

I am out of breath now...................Some typo may be there.........plzzzzz have some pity on me..........

Another method.....

(r²+r1²)+(r2²+r3²)

= 16R²sin²(A/2) [ sin²(B/2) . sin²(C/2) + cos²(B/2) . cos²(C/2) ]
+
16R²cos²(A/2) [ sin²(B/2) . cos²(C/2) + cos²(B/2) . sin²(C/2) ]

= 16R²sin²(A/2) [ {(1- cos B)/2} . {(1- cos c)/2} +
{(1+ cos B)/2} . {(1+ cos C)/2} ]
+
16R²cos²(A/2) [ {(1- cos B)/2} . {(1+ cos c)/2} +
{(1+ cos B)/2} . {(1- cos C)/2} ]

= 4R²sin²(A/2) [ 2 + 2 cos B . cos C ]
+
4R²cos²(A/2) [ 2 - 2 cos B . cos C ]

= 8R² + 4R² [ {cos (B+C) + cos (B-C)} . {sin²(A/2) - cos²(A/2)} ]

= 8R² - 4R² [ {cos A . (- cos A)} + {cos (B+C) . cos (B-C)} ]

= 8R² + 4R² [ cos²A + cos²B - sin²C ]

= 8R² + 4R² [ 2 - (sin²A + sin²B + sin²C) ]

= 16R² - (a² + b² + c²)

So,
(a² + b² + c² + r² + r1² + r2² + r3²) + R² = 16R² + R² = 17R²