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Where can I find the detailed answer to question 201kloc-0/23 of Shanghai senior high school entrance examination?
& ltp & gt23. As shown in the figure, in trapezoidal ABCD, ad∨BC, AB=DC, the intersection point D is DE⊥BC, and the vertical feet E and DE extend to F, so that EF = DE connects BF, CD and AC.

& ltp>( 1) Verification: the quadrilateral ABFC is a parallelogram; & lt/p & gt;

& ltp & gt(2) If DE2=BE? CE, prove it. The quadrilateral ABFC is a rectangle.

& ltp> solution:

& ltp> proves that:

& ltp>( 1) connects to BD,

& ltp>∫ ABCD, AD∨BC, AB=DC in the trapezoid,

& ltp & gt∴ac=bd,∠acb=∠dbc<; /p & gt;

& ltp & gt∵de⊥bc,ef=de,<; /p & gt;

& ltp & gt∴bd=bf,∠dbc=∠fbc,<; /p & gt;

& ltp & gt∴ac=bf,∠acb=∠cbf<; /p & gt;

& ltp & gt∴ac∥bf,<; /p & gt;

& ltp>∴ Quadrilateral ABFC is a parallelogram; & lt/p & gt;

& ltp & gt(2)∫DE2 = BE? CE & lt/p & gt;

& ltp>∴ Derby = Said,</p >

& ltp & gt∠∠DEB =∠DEC = 90,& lt/p & gt;

& ltp & gt∴△bde∽△dec,<; /p & gt;

& ltp & gt∴∠cde=∠dbe,<; /p & gt;

& ltp & gt∴∠bfc=∠bdc=∠bde+∠cde=∠bde+∠dbe=90,& lt/p & gt;

& ltp>∴ quadrilateral ABFC is a rectangle.

& ltp & gt& lt/p & gt;