例如图1,正方形ABCD的对角线AC与BD交于点O,以正方形的边BC为斜边在正方形内作直角三角形BCE,且∠BEC=90°,CE=3,BE=5,求S_(△OBE).解法1因为四边形ABCD是正方形,AC,BD是对角线,所以AC⊥BD,OB=OC,则∠BOC=90°,∠BCO=45°,因为∠BEC=90°,所以B,C,E,O四点共圆,则∠BEO=∠BCO=45°, For example, in Figure 1, the diagonal lines AC and BD of the square ABCD are intersected with the point O, the right triangle BCE is formed in the square with the side BC of the square, and BEC = 90 °, CE = 3 and BE = 5. S Δ OBE. Solution 1 Since the quadrilateral ABCD is a square and AC and BD are diagonal, ACBD and OB = OC, then ∠BOC = 90 ° and ∠BCO = 45 ° because ∠BEC = 90 ° , So B, C, E, O at four o’clock, then ∠BEO = ∠ BCO = 45 °,