以AD、BD为邻边作平行四边形ADBE.
∵ADBE是平行四边形,∴AE=BD=√13/2,AE∥BD,BE=AD=1,BE∥DA.
又AD⊥BC,∴BE⊥BC,∴由勾股定理,有:CE^2=BE^2+BC^2=1+(√3)^2=4.
由AE=√13/2、AC=√3/2、CE^2=4,得:AE^2+AC^2=CE^2,
∴由勾股定理的逆定理,有:AE⊥AC,结合AE∥BD,得:AC⊥BD,
∴AC与BD所成的角为90°.
以AD、BD为邻边作平行四边形ADBE.
∵ADBE是平行四边形,∴AE=BD=√13/2,AE∥BD,BE=AD=1,BE∥DA.
又AD⊥BC,∴BE⊥BC,∴由勾股定理,有:CE^2=BE^2+BC^2=1+(√3)^2=4.
由AE=√13/2、AC=√3/2、CE^2=4,得:AE^2+AC^2=CE^2,
∴由勾股定理的逆定理,有:AE⊥AC,结合AE∥BD,得:AC⊥BD,
∴AC与BD所成的角为90°.