Question Number 207362 by mr W last updated on 12/May/24
Answered by A5T last updated on 12/May/24
$${Ptolemy}'{s}\:{theorem}:\:{AC}×{BP}={AP}×{BC}+{AB}×{PC} \\ $$$${AB}={AC}={BC}\Rightarrow{PB}={PA}+{PC} \\ $$
Answered by sniper237 last updated on 13/May/24
![Let Q ∈ [BP] such as PQ=PC As mesBPC=mesBAC=60 then QPC is equilateral as ABC So r(A)=B and r(P)=Q with r=R(C;(π/3)) Thus r is an isometry, PA=BQ Then PB=PQ+QB=PC+PA](https://www.tinkutara.com/question/Q207373.png)
$${Let}\:{Q}\:\in\:\left[{BP}\right]\:{such}\:{as}\:{PQ}={PC} \\ $$$${As}\:{mesBPC}={mesBAC}=\mathrm{60}\:{then} \\ $$$${QPC}\:{is}\:{equilateral}\:{as}\:{ABC} \\ $$$${So}\:\:{r}\left({A}\right)={B}\:{and}\:{r}\left({P}\right)={Q}\:{with}\:{r}={R}\left({C};\frac{\pi}{\mathrm{3}}\right)\: \\ $$$${Thus}\:{r}\:{is}\:{an}\:{isometry},\:{PA}={BQ} \\ $$$${Then}\:{PB}={PQ}+{QB}={PC}+{PA} \\ $$