Question Number 194815 by universe last updated on 16/Jul/23
Answered by sniper237 last updated on 16/Jul/23
$${Incenter}\:{of}\:\:\angle{ABC}\:\:\:{is}\:\:{bar}\left\{\left({A},{a}\right),\left({B},{b}\right),\left({C},{c}\right)\right\} \\ $$$${meaning}\:\:{aI}\overset{\rightarrow} {{A}}+{bI}\overset{\rightarrow} {{B}}+{cI}\overset{\rightarrow} {{C}}=\overset{\rightarrow} {\mathrm{0}}\:\:\:\left(\ast\right) \\ $$$${hence}\:\:{aI}'\overset{\rightarrow} {{A}}'+{bI}'\overset{\rightarrow} {{B}}'+{cI}'\overset{\rightarrow} {{C}}'=\overset{\rightarrow} {\mathrm{0}}\:\:\left(\ast\ast\right)\:\:{by}\:{projecion} \\ $$$${Afer}\:{inroducing}\:{I}'\:{in}\:\left(\ast\right)\:{we}\:{have} \\ $$$$\left({a}+{b}+{c}\right){I}\overset{\rightarrow} {{I}}'\:=\:{a}\left({A}\overset{\rightarrow} {{A}}+{A}'\overset{\rightarrow} {{I}}'\right)\:+\:{b}\left({B}\overset{\rightarrow} {{B}}'+{B}'\overset{\rightarrow} {{I}}'\right)+{c}\left({C}\overset{\rightarrow} {{C}}'+{C}'\overset{\rightarrow} {{I}}'\right) \\ $$$$=\:{a}\:{A}\overset{\rightarrow} {{A}}'+{bB}\overset{\rightarrow} {{B}}'+{cC}\overset{\rightarrow} {{C}}'\:\:{applying}\:\left(\ast\ast\right) \\ $$$${so}\:\left({a}+{b}+{c}\right){II}'={a}.{a}'+{b}.{b}'+{c}.{c}'\:\:{since}\:{all}\:{following}\:{same}\:{direcion} \\ $$