Question Number 84718 by Power last updated on 15/Mar/20
Answered by mr W last updated on 15/Mar/20
$${AC}^{\mathrm{2}} ={PA}^{\mathrm{2}} +{PC}^{\mathrm{2}} −\mathrm{2}×{PA}×{PC}\:\mathrm{cos}\:\mathrm{120}° \\ $$$$\Rightarrow{AC}^{\mathrm{2}} ={PA}^{\mathrm{2}} +{PC}^{\mathrm{2}} +{PA}×{PC} \\ $$$$\frac{{PA}^{\mathrm{2}} +{PB}^{\mathrm{2}} −{AB}^{\mathrm{2}} }{\mathrm{2}×{PA}×{PB}}=\frac{{PC}^{\mathrm{2}} +{PB}^{\mathrm{2}} −{BC}^{\mathrm{2}} }{\mathrm{2}×{PC}×{PB}} \\ $$$$\frac{{PA}^{\mathrm{2}} +{PB}^{\mathrm{2}} −{PA}^{\mathrm{2}} −{PC}^{\mathrm{2}} −{PA}×{PC}}{{PA}}=\frac{{PC}^{\mathrm{2}} +{PB}^{\mathrm{2}} −{PA}^{\mathrm{2}} −{PC}^{\mathrm{2}} −{PA}×{PC}}{{PC}} \\ $$$$\frac{{PB}^{\mathrm{2}} −{PC}^{\mathrm{2}} −{PA}×{PC}}{{PA}}=\frac{{PB}^{\mathrm{2}} −{PA}^{\mathrm{2}} −{PA}×{PC}}{{PC}} \\ $$$${PB}^{\mathrm{2}} ×{PC}−{PC}^{\mathrm{3}} −{PA}×{PC}^{\mathrm{2}} ={PB}^{\mathrm{2}} ×{PA}−{PA}^{\mathrm{3}} −{PA}^{\mathrm{2}} ×{PC} \\ $$$${PB}^{\mathrm{2}} ×\left({PC}−{PA}\right)={PC}^{\mathrm{3}} −{PA}^{\mathrm{3}} +{PA}×{PC}\left({PC}−{PA}\right) \\ $$$${PB}^{\mathrm{2}} ={PC}^{\mathrm{2}} +{PA}×{PC}+{PA}^{\mathrm{2}} +{PA}×{PC} \\ $$$${PB}^{\mathrm{2}} =\left({PA}+{PC}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{PB}={PA}+{PC} \\ $$
Commented by Power last updated on 15/Mar/20
$$\mathrm{thanks} \\ $$
Answered by behi83417@gmail.com last updated on 15/Mar/20
$$\mathrm{PA}.\mathrm{BC}+\mathrm{PC}.\mathrm{AB}=\mathrm{PB}.\mathrm{AC}\overset{\mathrm{BA}=\mathrm{BC}=\mathrm{AC}} {\Rightarrow} \\ $$$$\left(\mathrm{PA}+\mathrm{PC}\right).\mathrm{BC}=\mathrm{PB}.\mathrm{AC} \\ $$$$\mathrm{PA}+\mathrm{PC}=\mathrm{PB}\:\:\left(\mathrm{by}\:\mathrm{using}\:\mathrm{petolemy}'\mathrm{s}\:\mathrm{teorem}\right) \\ $$
Commented by mr W last updated on 15/Mar/20
$${nice}\:{way}\:{sir}! \\ $$