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$\\a)\ Solution:\\\\ \frac{(b+c)}{a},\ \frac{(c+a)}{b}\ and\ \frac{(a+b)}{c}\ are\ in \ AP\\\\\\ Considering\ the\ common\ difference\ of\ terms\\\\ \Rightarrow \frac{(c+a)}{b}-\frac{(b+c)}{a}=\frac{(a+b)}{c}-\frac{(c+a)}{b}\\\\\\ \Rightarrow \frac{a(c+a)-b(b+c)}{ab}=\frac{b(a+b)-c(c+a)}{bc}\\\\\\ \Rightarrow \frac{ac+a^2-b^2-bc}{ab}=\frac{ab+b^2-c^2-ac}{bc}\\\\\\ \Rightarrow \frac{c(a-b)+(a+b)(a-b)}{ab}=\frac{a(b-c)+(b+c)(b-c)}{bc}\\\\\\ \Rightarrow \frac{(a-b)(c+a+b)}{ab}=\frac{(b-c)(a+b+c))}{bc}\\\\\\ \Rightarrow \frac{a-b}{ab}=\frac{b-c}{bc}\\\\\\ \Rightarrow \frac{a}{ab}-\frac{b}{ab}=\frac{b}{bc}-\frac{c}{bc}\\\\\\ \Rightarrow \frac1b-\frac1a=\frac1c-\frac1b\\\\\\ \Rightarrow \frac1b+\frac1b=\frac1a+\frac1c$

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$\\Solution:\\\\ Given:\\\\ a^2+b^2+c^2-ab-bc-ca=0\\\\ \Rightarrow 2(a^2+b^2+c^2-ab-bc-ca)=0\\\\ \Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\\\ \Rightarrow (a^2+b^2-2ab)+(b^2+c^2-2bc)+(c^2+a^2-2ca)=0\\\\ \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\\\\ Since\ sum\ of\ the\ squares\ are\ zero\ then\\\\ \Rightarrow (a-b)^2=0\Rightarrow a=b\\\\ \Rightarrow (b-c)^2=0\Rightarrow b=c\\\\ \Rightarrow (c-a)^2=0\Rightarrow c=a\\\\ \therefore a=b=c$

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$\\Solution:\\ Given:\\From\ figure\\ AQ\ and\ AR\ are\ the\ tangents\ then,\ AQ=AR\\ BQ, BP\ and\ CP,CR\ are\ the\ tangents\ then,\\ BQ=BP\ and\\ CR=CP \\\\RTP:\\ AQ=\frac{1}{2}(AB+BC+AC)\\\\ Proof:\\ From\ figure:\\ AQ=AB+BQ\\ AQ=AB+BP\ \ \ \ \rightarrow (1)\ \ \ [\because From\ given]\\ \\AR= AC+CR\\ AR=AC+PC\ \ \ \ \rightarrow (2)\ \ \ [\because From\ given]\\ \\Add\ equations\ (1)\ and\ (2)\\\\ \Rightarrow AQ+AR=AB+BP+AC+PC\\\\ \Rightarrow 2AQ=AB+BC+AC\ \ \ \ \ \ [\because from\ given]\\\\ \therefore AQ=\frac{1}{2}(AB+BC+AC)\\\\ Hence\ proved.$

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