in triangle ABC if b+c/11 = c+a/12 = a+b/13

PT cosA/7 = cosB/19 = cosC/25

Given, b+c11=c+a12=a+b13=ksaySo b+c=11k.....1c+a=12k......2a+b=13k..........3Adding all the equation we get, 2a+b+c=36kor a+b+c=18kNow subtracting equation 1, 2, 3 from the equation one by one we will get, a=7k, b = 6k , c=5kNow using cosine formulas, cosA=b2+c2-a22bc=6k2+5k2-7k22×6k×5k=15=735or cosA7=135........4cosB=a2+c2-b22ac=7k2+5k2-6k22×7k×5k=1935cosB19=135.......5cosC=b2+a2-c22ba=6k2+7k2-5k22×6k×7k=57=2535cosC25=135........6Now using equation 4, 5, 6 we get, cosA7=cosB19=cosC25Hence proved.

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