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$\frac{2E}{a}=\frac{2E}{b}+\frac{2E}{c}=>a=\frac{bc}{b+c}$$a^{2}+b^{2}+c^{2}=\frac{b^{2}c^{2}}{\left ( b+c \right )^{2}}+b^{2}+c^{2}$=$\frac{\left ( bc+b^{2}+c^{2} \right )^{2}}{\left ( b+c \right )^{2}}$=$(b+c-a)^{2}$
$\frac{2E}{a}=\frac{2E}{b}+\frac{2E}{c}=>a=\frac{bc}{b+c}$
ΑπάντησηΔιαγραφή$a^{2}+b^{2}+c^{2}=\frac{b^{2}c^{2}}{\left ( b+c \right )^{2}}+b^{2}+c^{2}$=$\frac{\left ( bc+b^{2}+c^{2} \right )^{2}}{\left ( b+c \right )^{2}}$=$(b+c-a)^{2}$