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Question Number 84718 by Power last updated on 15/Mar/20

	Answered by mr W last updated on 15/Mar/20

	${A C}^{2} = {P A}^{2} + {P C}^{2} - 2 \times P A \times P C \cos 120 °$ $\Rightarrow {A C}^{2} = {P A}^{2} + {P C}^{2} + P A \times P C$ $\frac{{P A}^{2} + {P B}^{2} - {A B}^{2}}{2 \times P A \times P B} = \frac{{P C}^{2} + {P B}^{2} - {B C}^{2}}{2 \times P C \times P B}$ $\frac{{P A}^{2} + {P B}^{2} - {P A}^{2} - {P C}^{2} - P A \times P C}{P A} = \frac{{P C}^{2} + {P B}^{2} - {P A}^{2} - {P C}^{2} - P A \times P C}{P C}$ $\frac{{P B}^{2} - {P C}^{2} - P A \times P C}{P A} = \frac{{P B}^{2} - {P A}^{2} - P A \times P C}{P C}$ ${P B}^{2} \times P C - {P C}^{3} - P A \times {P C}^{2} = {P B}^{2} \times P A - {P A}^{3} - {P A}^{2} \times P C$ ${P B}^{2} \times (P C - P A) = {P C}^{3} - {P A}^{3} + P A \times P C (P C - P A)$ ${P B}^{2} = {P C}^{2} + P A \times P C + {P A}^{2} + P A \times P C$ ${P B}^{2} = {(P A + P C)}^{2}$ $\Rightarrow P B = P A + P C$
	Commented by Power last updated on 15/Mar/20

	$thanks$
	Answered by behi83417@gmail.com last updated on 15/Mar/20

	$PA . BC + PC . AB = PB . AC \overset{BA = BC = AC}{\Rightarrow}$ $(PA + PC) . BC = PB . AC$ $PA + PC = PB (by using {petolemy}^{'} s teorem)$
	Commented by mr W last updated on 15/Mar/20

	$n i c e w a y s i r!$
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