& 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;