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Prove that `r_1+r_2+r_3-r=4R` |
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Answer» `r_(1) + r_(2) + r_(3) - r = (Delta)/(s -a) + (Delta)/(s -b) + (Delta)/(s-c) -(Delta)/(s)` `= Delta [(s -b + s-a)/((s -a) (s-b)) + ((sis + c))/(s(s -c))]` `= Delta [(c)/((s-a)(s-b)) + (c)/(s(s -c))]` `= Deltac [(s(s-c) + (s -a) (s-b))/(s(s -a) (s-b) (s-c))]` `= (Deltac)/(Delta^(2)) [2s^(2) -s (a + b + c) + ab]` `= ((c)/(Delta)) [2s^(2) -s (2s) + ab]` `= 4 ((abc)/(4Delta)) = 4R` |
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