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hbghlyj
Posted 2019-8-1 01:46
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(1)设正方形边长为2,$\therefore {\text{PM}} = {\text{PF}},{\text{QG}} = {\text{QM}}$,在${\text{Rt}}\vartriangle {\text{PCQ}}$中,${\text{P}}{{\text{Q}}^2} = {\text{P}}{{\text{C}}^2} + {\text{C}}{{\text{Q}}^2},\therefore {\left( {{\text{PF}} + {\text{QG}}} \right)^2} = {\left( {1 - {\text{PF}}} \right)^2} + {\left( {1 - QG} \right)^2},\therefore PF\cdot{\text{QG}} = 1 - {\text{PF}} - {\text{QG}},{\text{BP}}\cdot{\text{DQ}} = \left( {1 + {\text{PF}}} \right)\left( {1 + {\text{QG}}} \right) = 2 = {\text{BE}}\cdot{\text{AD}},\vartriangle {\text{ADQ}} \sim \vartriangle {\text{PBE}},{\text{AP}}\parallel {\text{HQ}}$ |
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