Category: Geometry

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…

Question Number 207230 by mr W last updated on 10/May/24 Answered by mr W last updated on 10/May/24 Commented by mr W last updated on…

Question Number 207035 by mr W last updated on 04/May/24 Answered by A5T last updated on 04/May/24 $${BE}=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{10}}\Rightarrow{EF}={FB}=\sqrt{\mathrm{10}} \\ $$$${Let}\:{the}\:{line}\:{through}\:{F}\:{parallel}\:{to}\:{BC}\:{meet}\:{AB},{DC} \\ $$$${at}\:{H},{I}\:{resp}.,{then}\:{BH}=\mathrm{3}={DI};\:{FH}=\mathrm{1} \\…

Question Number 207008 by mr W last updated on 03/May/24 Answered by A5T last updated on 03/May/24 Commented by A5T last updated on 03/May/24 $$\frac{\mathrm{5}}{{BO}}=\frac{\mathrm{6}}{{CO}}\Rightarrow\frac{{BO}}{{CO}}=\frac{\mathrm{5}}{\mathrm{6}};\:{BO}=\mathrm{5}{x};{CO}=\mathrm{6}{x}…

Question Number 206937 by efronzo1 last updated on 01/May/24 Answered by A5T last updated on 01/May/24 Commented by A5T last updated on 01/May/24 $$\mathrm{8}×\mathrm{14}={x}\left({x}+\mathrm{2}{r}\right)={x}^{\mathrm{2}} +\mathrm{2}{rx}=\mathrm{112}…