Question-84718

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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